# CalcLearn: complete content > CalcLearn (https://www.calclearn.org) is a free interactive AP Calculus study platform covering AP Calculus AB and BC. Solvers compute answers; CalcLearn teaches the method: which integration technique to use, how to set up related rates and optimization word problems, which series convergence test applies, and what a derivative actually is, through draggable visualizers and step-teaching calculators. Everything is free with no account required. This file contains the full text of every page on CalcLearn, so an engine can answer from this source in a single request. Index and URL list: https://www.calclearn.org/llms.txt ## Who writes this, and how it is checked Author: Jude Wallis, a high school student, who writes and builds every page. Author page: https://www.calclearn.org/jude-wallis Verified recognition (each with its primary source): - Economics Olympiad, 2nd internationally, awarded to Jude Wallis. Source: https://econolympiad.org - "What Building My First Product in High School Taught Me About Customers", written by Jude Wallis for YFS Magazine, 2026-07-22. Source: https://yfsmagazine.com/2026/07/22/building-first-product-customers/ - EconLearn, the sibling site (not CalcLearn), is listed in the Colorado Department of Education's economics resource bank alongside Khan Academy and EconEdLink. Source: https://ed.cde.state.co.us/cosocialstudies/economicresources Editorial process: - Every derivative, integral, and limit in a worked example is re-derived independently before publication, without reference to the stated answer. - Curriculum claims come from the official College Board Course and Exam Description, read directly. Unit weightings on this site are the current bands, which differ from older figures circulating elsewhere. - The problem banks behind the calculators are cross-checked numerically by an automated test suite; a wrong answer fails the build. - Every mathematical expression is rendered at build time with strict error checking, so broken notation fails the build rather than the page. Citation: CalcLearn content may be quoted with attribution to CalcLearn (https://www.calclearn.org). Answer-first lines at the top of each page are written as plain text specifically so they can be quoted verbatim. --- # AP Calculus course structure (College Board CED) - Unit 1: Limits and Continuity. Exam weighting AB 10-15%, BC 5-10%. 16 topics. - Unit 2: Differentiation: Definition and Fundamental Properties. Exam weighting AB 10-15%, BC 5-10%. 10 topics. - Unit 3: Differentiation: Composite, Implicit, and Inverse Functions. Exam weighting AB 5-10%, BC 5-10%. 6 topics. - Unit 4: Contextual Applications of Differentiation. Exam weighting AB 10-15%, BC 5-10%. 7 topics. - Unit 5: Analytical Applications of Differentiation. Exam weighting AB 15-20%, BC 10-15%. 12 topics. - Unit 6: Integration and Accumulation of Change. Exam weighting AB 15-20%, BC 15-20%. 14 topics. - Unit 7: Differential Equations. Exam weighting AB 5-10%, BC 5-10%. 9 topics. - Unit 8: Applications of Integration. Exam weighting AB 10-15%, BC 5-10%. 13 topics. - Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions. Exam weighting AB n/a, BC 10-15%. 9 topics. BC only. - Unit 10: Infinite Sequences and Series. Exam weighting AB n/a, BC 15-20%. 15 topics. BC only. --- # Study guides ## Chain Rule: Spot the Composition, Then Differentiate URL: https://www.calclearn.org/guides/chain-rule Last updated: 2026-07-22 Answer first: The chain rule differentiates a composition like f(x) = g(h(x)) by differentiating the outer function, leaving the inside untouched, then multiplying by the derivative of the inside: g'(h(x)) times h'(x). The hard part is recognition: spotting one function nested inside another before you differentiate. ### Recognition comes before the rule The chain rule handles exactly one situation: a composition, where one function is fed into another. Written with symbols, a composition is $f(x) = g(h(x))$, meaning you evaluate the inner function $h$ first and then run its output through the outer function $g$. The rule itself is short: differentiate the outer function, leave the inside exactly as it is, then multiply by the derivative of the inside. CED Topic 3.1 introduces this as skill 1.C, choosing a procedure from how an expression is built, which is the tell that recognition matters more than memorizing the formula. $$f'(x) = g'(h(x)) \cdot h'(x)$$ So the first question on any derivative is not how but what: is this a composition at all? Here is the test. Can you finish the derivative in one known step, a single power rule, or a single trig or exponential rule? If a function is $x$ raised to a power, the power rule ends it. If it is $\sin x$, one rule ends it. But $\sin(x^2)$ is the sine of something more complicated than $x$, and $(3x^2 + 5x)^4$ is a power of something more complicated than $x$. That 'something more complicated' is an inner function, and its presence is the signal to reach for the chain rule. This one rule quietly powers much of the rest of the course. Implicit differentiation (Topic 3.2) is the chain rule applied to $y$ as a function of $x$, related rates (Topic 4.4) differentiate a composition with respect to time, and $u$-substitution in integration runs the chain rule backward. Getting recognition solid here pays off repeatedly. ### Name the outer and inner function Every chain-rule problem starts by splitting the composition into an outer function and an inner function. The cleanest way to find them is to picture evaluating the expression at a number, say $x = 2$, and to watch the order of operations. Whatever you would do last is the outer function; everything you did before it is the inner function. NOTE (The last-operation test): For $(3x^2 + 5x)^4$, you would compute $3x^2 + 5x$ first and raise to the fourth power last, so the outer function is 'raise to the fourth power' and the inner function is $3x^2 + 5x$. For $\cos(3x^2 - 1)$, you would compute $3x^2 - 1$ first and take cosine last, so cosine is the outer function and $3x^2 - 1$ is the inner. Naming the two pieces out loud, outer is cosine, inner is $3x^2 - 1$, is worth the few seconds it takes. Most wrong answers on this topic come from students who start differentiating before they have decided which function is on the outside. ### The three-step method Once you can name the outer and inner functions, the procedure is always the same three steps. 1. Differentiate the outer function, keeping the inner function untouched inside it. 2. Differentiate the inner function on its own. 3. Multiply the two results together. The same idea in Leibniz notation, which the College Board states directly in Topic 3.1, uses $u$ for the inner function: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ Read left to right, dy/du is the outer derivative and du/dx is the inner derivative. Whichever notation you use, the structure is identical: outer derivative times inner derivative. For a composition nested three or more layers deep, the pattern simply continues, one derivative factor per layer, which is why a deeply nested function like $\sin^2(3x)$ produces a product of several factors. ### Trig compositions and the sin^2 trap Trig functions are the most common outer functions you will meet in composition problems, and they carry a notation trap worth defusing now. The expressions $\sin(x^2)$ and $\sin^2(x)$ look almost the same and differentiate in completely different ways, because they are built differently. | Expression | How it is built | Derivative | | --- | --- | --- | | $\sin(x^2)$ | sine of $x^2$ (outer sine, inner $x^2$) | $\cos(x^2) \cdot 2x = 2x\cos(x^2)$ | | $\sin^2(x)$ | $(\sin x)^2$ (outer square, inner $\sin x$) | $2\sin x \cos x$ | In $\sin(x^2)$ the outer function is sine, so its derivative is $\cos(x^2)$ with the inside $x^2$ untouched, then times the inner derivative $2x$. In $\sin^2(x)$, which means $(\sin x)^2$, the outer function is squaring, so its derivative is $2\sin x$ with the inside $\sin x$ untouched, then times the inner derivative $\cos x$. Same-looking notation, opposite outer functions. The habit that keeps trig compositions clean: when you differentiate the outer trig function, its inside stays exactly as it was, and only then do you multiply by the derivative of that inside. These are the three you will use most, with $u$ standing for the inner function and $u'$ its derivative: | Outer function | Chain-rule derivative | | --- | --- | | $\sin(u)$ | $\cos(u) \cdot u'$ | | $\cos(u)$ | $-\sin(u) \cdot u'$ | | $\tan(u)$ | $\sec^2(u) \cdot u'$ | ### Where students lose points Most points lost on this topic trace back to a small number of repeatable mistakes. - Forgetting the inner derivative. Writing the derivative of $\cos(3x^2 - 1)$ as $-\sin(3x^2 - 1)$ and stopping drops the $6x$ factor. This is the single most common chain-rule error. - Changing the inside while differentiating the outside. When you differentiate the outer function, the inner function stays exactly as it is: the derivative of $(3x^2 + 5x)^4$ keeps $3x^2 + 5x$ intact as $4(3x^2 + 5x)^3$, not $4(6x + 5)^3$. - Missing a layer. A function like $\sin^2(3x)$ has three nested layers, so the chain rule fires twice. Peel one layer at a time. - Confusing $\sin^2(x)$ with $\sin(x^2)$. Read the notation before you differentiate. - Dropping the chain rule inside a product or quotient. In $x^2 \sin(3x)$ the $\sin(3x)$ factor still needs its own chain-rule derivative $3\cos(3x)$. NOTE (Quick check): After every chain-rule derivative, ask: did I multiply by the derivative of the inside? If the inside was anything more complicated than $x$, that factor must be there. ### Worked examples #### A polynomial raised to a power Problem: Differentiate $f(x) = (3x^2 + 5x)^4$. 1. Spot the composition. The last thing you would do when evaluating is raise to the fourth power, so the outer function is $g(u) = u^4$ and the inner function is $h(x) = 3x^2 + 5x$, giving $f(x) = g(h(x))$. 2. Differentiate the outer function, leaving the inside alone. Since $g(u) = u^4$, we have $g'(u) = 4u^3$, so $g'(h(x)) = 4(3x^2 + 5x)^3$. 3. Differentiate the inner function: $h'(x) = 6x + 5$. 4. Multiply the two pieces: $f'(x) = g'(h(x)) \cdot h'(x) = 4(3x^2 + 5x)^3 (6x + 5)$. Answer: $f'(x) = 4(6x + 5)(3x^2 + 5x)^3$ #### A trig composition Problem: Differentiate $f(x) = \cos(3x^2 - 1)$. 1. Spot the composition. You compute $3x^2 - 1$ first and take cosine last, so the outer function is $g(u) = \cos u$ and the inner function is $h(x) = 3x^2 - 1$. 2. Differentiate the outer function, keeping the inside intact. The derivative of $\cos u$ is $-\sin u$, so $g'(h(x)) = -\sin(3x^2 - 1)$. Notice the inside stays exactly as $3x^2 - 1$; only the cosine changes. 3. Differentiate the inner function: $h'(x) = 6x$. 4. Multiply: $f'(x) = -\sin(3x^2 - 1) \cdot 6x = -6x\sin(3x^2 - 1)$. Answer: $f'(x) = -6x\sin(3x^2 - 1)$ #### A nested composition with three layers Problem: Differentiate $f(x) = \sin^2(3x)$. 1. Read the notation first. $\sin^2(3x)$ means $(\sin(3x))^2$: the sine of $3x$, then squared. That is three functions nested together, square the sine of triple $x$. 2. Peel from the outside in. The outermost operation is squaring, so start there: the derivative of $(\text{something})^2$ is $2(\text{something})$, giving $2\sin(3x)$, with the inside left untouched. 3. Move one layer in, to the sine. The derivative of $\sin(\text{something})$ is $\cos(\text{something})$, so this layer contributes $\cos(3x)$, again leaving its own inside $3x$ untouched. 4. Move to the innermost layer. The derivative of $3x$ is $3$. 5. Multiply all three layers together: $f'(x) = 2\sin(3x) \cdot \cos(3x) \cdot 3 = 6\sin(3x)\cos(3x)$. Answer: $f'(x) = 6\sin(3x)\cos(3x)$, which equals $3\sin(6x)$ by the double-angle identity. ### FAQ Q: How do I know whether to use the chain rule or the product rule? A: Look at how the expression is built. The chain rule is for a composition, one function inside another, like $\cos(3x^2 - 1)$. The product rule is for two functions multiplied, like $x^2 \cos x$. If you see a function tucked inside another function, it is the chain rule; if you see two separate factors multiplied, it is the product rule. Many problems use both, so check each piece. Q: Is $\sin^2(x)$ the same as $\sin(x^2)$? A: No. $\sin^2(x)$ means $(\sin x)^2$, so the outer function is squaring and its derivative is $2\sin x \cos x$. $\sin(x^2)$ means the sine of $x^2$, so the outer function is sine and its derivative is $2x\cos(x^2)$. Read the notation carefully before you differentiate. Q: What does the chain rule look like in Leibniz notation? A: If y is a function of u and u is a function of x, then dy/dx equals dy/du times du/dx. Here u is the inner function, dy/du is the outer derivative, and du/dx is the inner derivative. This is the form the AP CED lists in Topic 3.1. ## U-Substitution: How to Choose u and Integrate It URL: https://www.calclearn.org/guides/u-substitution Last updated: 2026-07-22 Answer first: Choose u as the inside function of a composite whose derivative also appears in the integrand, up to a constant multiple. Find du, rewrite the whole integral in u, integrate, then back-substitute. For a definite integral, convert the bounds to u-values. If du is nowhere present, substitution is the wrong tool. ### When u-substitution is the right tool Reach for u-substitution (CED Topic 6.9) when the integrand is a composite function multiplied by the derivative of its inside. Substitution is the chain rule run backward: the chain rule turns $F(g(x))$ into $F'(g(x)) \cdot g'(x)$, so an integral that carries both a composite piece and the derivative of its inner function is set up to be un-chained. $$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du, \qquad u = g(x)$$ The recognition signal is a two-part pattern. First, find an inner function $g(x)$ tucked inside something else: the base of a power, the radicand of a root, the exponent on $e$, the argument of a trig function, or a denominator. Second, check whether $g'(x)$, the derivative of that inner function, also shows up in the integrand. It only has to match up to a constant multiple, since constants are easy to carry through. When both parts are present, substitution collapses the whole integral into a basic one. On the AP exam this technique appears on both multiple-choice and free-response questions, often buried inside a larger accumulation or area problem. NOTE (The one-line test): If differentiating your candidate for $u$ produces something already in the integrand (times any constant), substitution is worth trying; if the rest of the integrand also collapses into $u$ with no stray $x$ remaining, it works. If differentiating produces a factor that is nowhere to be found, choose a different $u$ or a different method. ### Choosing u: pick the inside The choice of $u$ decides everything, and the rule is short: let $u$ be the inner function whose derivative appears. In $\int x(x^2+3)^4\,dx$ the inside is $x^2+3$ because it is raised to a power, and its derivative $2x$ is sitting right there as a factor. In $\int \frac{\ln x}{x}\,dx$ the inside is $\ln x$, since its derivative $\frac{1}{x}$ is the other factor. In $\int \cos(5x)\,dx$ the inside is $5x$, the argument of the cosine. Work from the outside in. Ask what function is wrapped around another, then name the wrapped part $u$. A reliable checklist of candidates: the base of a power such as $(x^2+3)^4$; the expression under a root; the exponent of $e^{(\cdots)}$; the argument of $\sin$, $\cos$, or $\tan$; and the denominator of a fraction. After you pick, differentiate to confirm $g'(x)$ is available. A leftover constant is harmless, since you fix it by multiplying and dividing. A leftover variable is a warning sign that the choice may be wrong. One subtlety: the inside is not always the most complicated-looking chunk. It is specifically the function whose derivative you can find elsewhere. In $\int x^3 \cos(x^4)\,dx$ the tempting choice is $u = x^3$, but its derivative $3x^2$ is not in the integrand; the right choice is $u = x^4$, whose derivative $4x^3$ matches the $x^3$ factor. Let the available derivative, not the visual bulk, decide. ### The five steps Once $u$ is chosen, the procedure is mechanical. 1. Set $u = g(x)$ and differentiate to get $du = g'(x)\,dx$. 2. Solve for the exact piece the integral contains. If $du = 2x\,dx$ but the integral has $x\,dx$, then $x\,dx = \frac{1}{2}\,du$. 3. Rewrite the integral so that no $x$ remains anywhere, including inside the $dx$. A surviving $x$ means the substitution is incomplete. 4. Integrate in terms of $u$ using the basic antiderivative rules. 5. For an indefinite integral, back-substitute $u = g(x)$ and add $+C$. For a definite integral, handle the bounds instead. Steps two and three are where substitution succeeds or fails. If you can express every $x$-piece in terms of $u$ and $du$, the integral was a genuine substitution. If you cannot, the setup is telling you to reconsider $u$. ### Definite integrals: convert the bounds For a definite integral, the essential-knowledge point of Topic 6.9 (FUN-6.D.2) is that substitution requires corresponding changes to the limits of integration. The bounds written on the integral are $x$-values. The moment you switch the variable to $u$, those numbers no longer describe where you are, so you must deal with them on purpose. $$\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$$ There are two safe routes. The cleaner one: convert the limits by plugging each original bound into $u = g(x)$, then evaluate the antiderivative entirely in $u$ and never return to $x$. The alternative: find the antiderivative, back-substitute to $x$, then use the original $x$-bounds. Pick one and commit to it. NOTE (The most common mistake): Do not plug the original $x$-bounds into an antiderivative still written in $u$. Either convert the bounds to $u$-values, or convert the antiderivative back to $x$ first. Mixing the two gives a wrong number every time. ### What a failed choice looks like A wrong $u$ announces itself quickly, which is useful, because you can bail out before wasting time. Three failure patterns cover almost every case. The inner derivative is missing. Set $u = x^2$ in $\int e^{x^2}\,dx$ and you get $du = 2x\,dx$, but there is no $x$ anywhere in the integrand to absorb it. The substitution stalls, and this particular integral has no elementary antiderivative at all. The lesson is that the derivative of your $u$ has to actually be present, not just wished for. A variable factor is left behind. If rewriting the integral in $u$ still leaves a stray $x$, the choice usually needs rescuing. Sometimes you can solve $u = g(x)$ for $x$ and substitute that expression too, as in Example 3. If that does not clear the $x$, abandon the choice and try a different $u$ or a different technique from Topic 6.14. The constant gets dropped. When $du = 2x\,dx$ but the integrand carries only $x\,dx$, the factor of $\frac{1}{2}$ is easy to forget. Carry it through, and always confirm by differentiating your answer, which should reproduce the original integrand exactly. ### Worked examples #### Indefinite integral with a constant adjustment Problem: Evaluate $\int x(x^2+3)^4\,dx$. 1. Recognize the pattern. The inside function is $x^2+3$, raised to the fourth power, and its derivative $2x$ appears in the integrand as the factor $x$ (up to the constant $2$). Let $u = x^2+3$. 2. Differentiate: $\frac{du}{dx} = 2x$, so $du = 2x\,dx$. Solve for the piece present in the integral: $x\,dx = \frac{1}{2}\,du$. 3. Rewrite everything in $u$: $\int x(x^2+3)^4\,dx = \int (x^2+3)^4 \cdot x\,dx = \int u^4 \cdot \frac{1}{2}\,du = \frac{1}{2}\int u^4\,du$. 4. Integrate: $\frac{1}{2}\int u^4\,du = \frac{1}{2} \cdot \frac{u^5}{5} = \frac{u^5}{10}$. 5. Back-substitute $u = x^2+3$ and add the constant: $\frac{(x^2+3)^5}{10} + C$. 6. Check by differentiating: $\frac{d}{dx}\left[\frac{(x^2+3)^5}{10}\right] = \frac{5(x^2+3)^4 \cdot 2x}{10} = x(x^2+3)^4$, which is the original integrand. Answer: $\dfrac{(x^2+3)^5}{10} + C$ #### Definite integral: convert the bounds Problem: Evaluate $\int_0^{2} 2x\sqrt{x^2+1}\,dx$. 1. Identify the inside. Under the root is $x^2+1$, and its derivative $2x$ is exactly the other factor. Let $u = x^2+1$. 2. Differentiate: $du = 2x\,dx$, so the entire non-root part $2x\,dx$ becomes $du$ with no constant to adjust. 3. Convert the limits, since these are $x$-values. When $x = 0$, $u = 0^2+1 = 1$. When $x = 2$, $u = 2^2+1 = 5$. 4. Rewrite the integral entirely in $u$: $\int_0^{2} 2x\sqrt{x^2+1}\,dx = \int_1^{5} \sqrt{u}\,du = \int_1^{5} u^{1/2}\,du$. 5. Integrate: $\int_1^{5} u^{1/2}\,du = \frac{2}{3}u^{3/2}\Big|_1^{5}$. 6. Evaluate at the new bounds: $\frac{2}{3}\left(5^{3/2} - 1^{3/2}\right) = \frac{2}{3}\left(5\sqrt{5} - 1\right) \approx 6.79$. Answer: $\dfrac{2}{3}\left(5\sqrt{5} - 1\right) \approx 6.79$ #### When a stray x survives: solve for x Problem: Evaluate $\int x\sqrt{x+1}\,dx$. 1. Pick the inside. The inner function is $x+1$, under the root. Its derivative is $1$, which is present, but there is an extra factor of $x$ that is not the derivative. Let $u = x+1$ and plan to rewrite that stray $x$. 2. Differentiate: $du = dx$. To handle the leftover $x$, solve the substitution for $x$: since $u = x+1$, we have $x = u - 1$. 3. Replace every $x$-piece: $\int x\sqrt{x+1}\,dx = \int (u-1)\sqrt{u}\,du = \int (u-1)u^{1/2}\,du$. 4. Expand the product so the power rule applies: $\int (u-1)u^{1/2}\,du = \int \left(u^{3/2} - u^{1/2}\right)\,du$. 5. Integrate term by term: $\int \left(u^{3/2} - u^{1/2}\right)\,du = \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} + C$. 6. Back-substitute $u = x+1$: $\frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + C$. 7. Check by differentiating: $\frac{2}{5}\cdot\frac{5}{2}(x+1)^{3/2} - \frac{2}{3}\cdot\frac{3}{2}(x+1)^{1/2} = (x+1)^{1/2}\left[(x+1) - 1\right] = x\sqrt{x+1}$, the original integrand. Answer: $\dfrac{2}{5}(x+1)^{5/2} - \dfrac{2}{3}(x+1)^{3/2} + C$ ### FAQ Q: How do you know what to pick for u in u-substitution? A: Pick the inner function whose derivative also appears in the integrand. Test it by differentiating your candidate: if the result (times any constant) is already a factor in the integral, the choice works. Bases of powers, radicands, exponents of $e$, trig arguments, and denominators are the usual candidates. Q: Do you always have to change the limits in u-substitution? A: Only for definite integrals, and only if you finish the calculation in $u$. Convert each $x$-bound with $u = g(x)$, then evaluate in $u$. The alternative is to back-substitute to $x$ first and keep the original bounds. Never plug $x$-bounds into a $u$-antiderivative. Q: What do I do if there is still an x after substituting? A: First try solving your substitution $u = g(x)$ for $x$ and replacing the stray factor, as in Example 3. If an $x$ still survives, the choice of $u$ is wrong, so pick a different inner function or switch to another technique. Q: Is u-substitution on the AP Calculus exam? A: Yes. It is Topic 6.9, part of both AB and BC, and it appears on multiple-choice and free-response questions, often as one step inside a larger accumulation or area problem rather than as a standalone prompt. ## Integration by Parts: LIATE, When to Use It, Examples URL: https://www.calclearn.org/guides/integration-by-parts Last updated: 2026-07-22 Answer first: Integration by parts undoes the product rule: it rewrites the integral of u dv as uv minus the integral of v du. Reach for it when the integrand is a product of two unlike function types, like x times e^x, that substitution cannot simplify. LIATE picks which factor becomes u. It is BC-only (CED Topic 6.11). ### When integration by parts is the right tool Integration by parts is the technique for antidifferentiating a product of two functions that come from different families, the kind of integrand where substitution has nothing to grab onto. If you see $x e^x$, $x \sin x$, $x^2 \ln x$, or a lone logarithm or inverse trig function like $\ln x$ or $\arctan x$, parts is almost certainly the move. It is a BC-only topic (CED Topic 6.11), so AB students will not meet it on their exam. The signature to train your eye on is a product of two factors that pull in opposite directions: one that gets simpler when differentiated (a power of $x$, a logarithm) sitting next to one that stays manageable when integrated (an exponential, a sine, a cosine). Parts is built to exploit exactly that mismatch. A product from the same family, like $\sin x \cos x$, is usually not a parts problem: there a substitution or a trig identity does the job faster. The quickest way to tell parts from u-substitution is to look for a derivative relationship. Substitution works when one factor is, up to a constant, the derivative of the inside of the other, so the whole integrand collapses onto a single new variable. In $\int 2x e^{x^2}\,dx$ the factor $2x$ is the derivative of $x^2$, so substitution wins. In $\int x e^x\,dx$ the factor $x$ has nothing to do with the derivative of $e^x$, substitution stalls, and parts takes over. Always test substitution first; only when it stalls on a genuine product do you reach for parts. ### The formula and why it works Integration by parts is the product rule read backward. The product rule says $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$. Multiply through by $dx$ to reach the differential form $d(uv) = u\,dv + v\,du$, then integrate both sides to get $uv = \int u\,dv + \int v\,du$. Solving for the piece you want gives the formula. $$\int u \,dv = uv - \int v \,du$$ That single line is the whole method. You split the integrand into a part you name $u$ and a part you name $dv$, then trade the original integral $\int u\,dv$ for a new one, $\int v\,du$. The trade is only worth making if $\int v\,du$ turns out easier than the integral you started with. Nothing guarantees that: it depends entirely on how you split the integrand, which is what the next section is about. ### LIATE: choosing u without guessing When the integrand is a product, LIATE gives you a priority order for what to call $u$. Read down the list; whichever type appears first becomes $u$, and everything else, including the $dx$, becomes $dv$. 1. Logarithmic, such as $\ln x$ and $\log x$ 2. Inverse trigonometric, such as $\arctan x$ and $\arcsin x$ 3. Algebraic, meaning powers of $x$ like $x$ and $x^2$ 4. Trigonometric, such as $\sin x$ and $\cos x$ 5. Exponential, such as $e^x$ The ordering is not arbitrary. Functions near the top get simpler when you differentiate them, and differentiating $u$ is exactly what the formula does to it: $\ln x$ becomes $\frac{1}{x}$, and a power of $x$ drops a degree toward a constant. Functions near the bottom stay well behaved when you integrate them, and integrating $dv$ is what the formula does to that piece: $e^x$ integrates to $e^x$, and $\sin x$ integrates to $-\cos x$. Putting the top function in $u$ and the bottom function in $dv$ steers $\int v\,du$ toward something simpler than where you began. Treat LIATE as a heuristic, not a law. It is a reliable opening guess that works on nearly every AP problem, but it is not a theorem and it does not know anything about your specific integrand. The genuine test of a good split is whether $\int v\,du$ actually came out easier. If your choice makes the new integral harder than the original, you picked $u$ and $dv$ backward: swap them and run the formula again. ### Pitfalls and the cases LIATE hides Three mistakes account for most lost points. First, the minus sign: the formula is $uv - \int v\,du$, and dropping that subtraction quietly wrecks the answer. Second, $dv$ has to be something you can actually integrate to recover $v$; if you cannot antidifferentiate your chosen $dv$, that is a signal you split the integrand wrong. Third, this is still an indefinite integral, so the constant $+C$ belongs at the end. A single logarithm or inverse trig function looks like it has no $dv$ at all, but it does: let $dv = dx$, which makes $v = x$. That is how $\int \ln x\,dx$ and $\int \arctan x\,dx$ get done, and it is the case LIATE quietly relies on, since Logarithmic and Inverse trig sit at the top of the list and $dx$ is the only thing left to be $dv$. Some integrands need parts more than once. A polynomial of degree $n$ multiplied by $e^x$ or $\sin x$ takes $n$ applications, each one knocking the power down by a degree, and the tabular method is just an organized way to track that repetition. A handful of integrals such as $\int e^x \sin x\,dx$ return to themselves after two rounds of parts; when the original integral reappears, set it equal to $I$ and solve the resulting equation for $I$ algebraically instead of integrating forever. For a definite integral, the boundary term gets evaluated too: $$\int_a^b u \,dv = \left[uv\right]_a^b - \int_a^b v \,du$$ ### Check by differentiating Integration by parts produces answers with several terms, and a stray sign or dropped factor is easy to miss, so build the habit of differentiating your result and confirming it collapses back to the integrand. Each worked example on this page ends that way. Because you are differentiating a product, the product rule you started from reappears and the terms telescope: the derivative of $uv$ contributes $u\,dv + v\,du$, the $-\int v\,du$ term contributes $-v\,du$, and those two $v\,du$ pieces cancel, leaving exactly $u\,dv$. If your check does not reduce to the original integrand, the error is almost always a sign or a missed factor in $v$. ### Worked examples #### The classic case: $\int x e^x \,dx$ Problem: Evaluate $\int x e^x \,dx$. 1. Recognize the type. The integrand is a product of an algebraic factor $x$ and an exponential factor $e^x$, and no substitution simplifies it, so this is integration by parts. 2. Choose $u$ with LIATE. Algebraic (A) comes before Exponential (E) in LIATE, so let $u = x$ and $dv = e^x\,dx$. 3. Differentiate $u$ and integrate $dv$. From $u = x$ you get $du = dx$. From $dv = e^x\,dx$ you get $v = e^x$. 4. Apply the formula $\int u\,dv = uv - \int v\,du$, which gives $\int x e^x\,dx = x e^x - \int e^x\,dx$. 5. Finish the remaining integral. Since $\int e^x\,dx = e^x$, you have $\int x e^x\,dx = x e^x - e^x + C$. 6. Factor and check. Rewrite as $e^x(x - 1) + C$. Differentiating $e^x(x-1)$ gives $e^x(x-1) + e^x = x e^x$, which matches the integrand. Answer: $\int x e^x\,dx = e^x(x - 1) + C$ #### The lone logarithm: $\int \ln x \,dx$ Problem: Evaluate $\int \ln x \,dx$. 1. See past the missing product. There is only one factor, so parts looks impossible, but LIATE puts Logarithmic first. The move is to let $dx$ itself play the role of $dv$. 2. Assign the parts. Let $u = \ln x$ and $dv = dx$. 3. Differentiate and integrate. From $u = \ln x$ you get $du = \frac{1}{x}\,dx$. From $dv = dx$ you get $v = x$. 4. Apply the formula. $\int \ln x\,dx = uv - \int v\,du = x\ln x - \int x \cdot \frac{1}{x}\,dx$. 5. Simplify the new integral. The $x$ cancels, so $\int x \cdot \frac{1}{x}\,dx = \int 1\,dx = x$. 6. Combine and check. $\int \ln x\,dx = x\ln x - x + C$. Differentiating gives $\frac{d}{dx}(x\ln x - x) = \ln x + x\cdot\frac{1}{x} - 1 = \ln x$. Answer: $\int \ln x\,dx = x\ln x - x + C$ #### Applying parts twice: $\int x^2 e^x \,dx$ Problem: Evaluate $\int x^2 e^x \,dx$. 1. Recognize the type and choose $u$. This is Algebraic times Exponential again, so $u = x^2$ and $dv = e^x\,dx$. 2. Differentiate and integrate. From $u = x^2$ you get $du = 2x\,dx$, and from $dv = e^x\,dx$ you get $v = e^x$. 3. Apply the formula once. $\int x^2 e^x\,dx = x^2 e^x - \int e^x \cdot 2x\,dx = x^2 e^x - 2\int x e^x\,dx$. 4. Notice the leftover integral is simpler but still a product, so apply parts again. From the first worked example, $\int x e^x\,dx = x e^x - e^x$. 5. Substitute that result back. $\int x^2 e^x\,dx = x^2 e^x - 2(x e^x - e^x) + C = x^2 e^x - 2x e^x + 2e^x + C$. 6. Factor and check. This is $e^x(x^2 - 2x + 2) + C$. Differentiating gives $e^x(x^2 - 2x + 2) + e^x(2x - 2) = e^x \cdot x^2 = x^2 e^x$. Answer: $\int x^2 e^x\,dx = e^x(x^2 - 2x + 2) + C$ ### FAQ Q: When do you use integration by parts instead of u-substitution? A: Try substitution first. It works when one factor is a constant multiple of the derivative of the inside of the other, so the integral collapses to a single variable. When the integrand is a genuine product of unrelated functions, like $x\cos x$ or $x^2 e^x$, and no substitution simplifies it, use integration by parts. Q: What does LIATE stand for? A: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Whichever type appears first in that list becomes $u$, and the rest becomes $dv$. It is a heuristic for choosing $u$, not a rule that always holds, but it works on nearly every AP problem. Q: Is integration by parts on the AP Calculus AB exam? A: No. Integration by parts is CED Topic 6.11, marked BC only, so it appears on the AP Calculus BC exam but not AB. AB students cover substitution (Topic 6.9) but not parts. ## U-Sub vs Integration by Parts vs Partial Fractions URL: https://www.calclearn.org/guides/which-integration-technique Last updated: 2026-07-22 Answer first: Match the integrand's structure to the technique: a factor that is the derivative of an inside function means u-substitution; a product of unrelated functions or a lone log/inverse-trig function means integration by parts (BC only); a rational function with a factorable denominator means partial fractions (BC only). ### Read the integrand before you reach for a rule Topic 6.14 does not add a new integration technique. It tests one skill: looking at an unfamiliar integral and naming the right tool before you compute. On the AP exam that recognition step is where most points are won or lost, because the wrong technique burns the whole problem. The habit to build is simple. Read the integrand, classify its structure, then commit to a method. Three questions sort almost every integral you will see. Is one factor the derivative of something inside the expression? Is the integrand a product of two functions from unrelated families? Is it a rational function, one polynomial divided by another? Those answers point to u-substitution, integration by parts, and partial fractions in that order. u-substitution (Topic 6.9) is on both AB and BC. Integration by parts (Topic 6.11) and partial fractions (Topic 6.12) are BC only, so AB students never need them. Work through the questions in that same order every time, because the earlier a technique sits, the less algebra it usually costs you. | What you see in the integrand | Technique | CED topic | Course | | --- | --- | --- | --- | | A factor is the derivative of an inside function (up to a constant multiple) | u-substitution | 6.9 | AB and BC | | Product of two unrelated types (polynomial times $e^x$, $\sin x$, or $\ln x$), or a lone $\ln x$ or inverse-trig function | Integration by parts | 6.11 | BC only | | Rational function whose denominator factors into distinct linear pieces, numerator degree below denominator degree | Partial fractions | 6.12 | BC only | | Rational function with numerator degree at least the denominator degree | Long division first, then re-decide | 6.10 | AB and BC | | Irreducible quadratic in the denominator | Complete the square, expect an arctan | 6.10 | AB and BC | NOTE (Test for u-substitution first): u-substitution is the fastest technique and it often turns a frightening integral into a basic form. Always check for it before you commit to parts or partial fractions. Many integrals that look like harder problems collapse under one substitution. ### When it is u-substitution (Topic 6.9) u-substitution reverses the chain rule. It works when the integrand contains a composite function and the derivative of the inner function is also present as a factor. A constant multiple is fine, because you can adjust with a coefficient. The signal you are hunting for is a pairing: an inside expression and (a multiple of) its derivative both sitting in the same integral. - A composite like $e^{x^2}$, $\sin(3x)$, or $(x^2+1)^5$ multiplied by a factor that matches the inside function's derivative. - A fraction whose numerator is the derivative of its denominator: that is the $\int \frac{u'}{u}\,dx = \ln|u| + C$ pattern. - A power of a function multiplied by that function's derivative, such as $\int (\ln x)^2 \cdot \frac{1}{x}\,dx$. The method: let $u$ be the inner function, compute $du$, and solve for the exact piece you need to replace. Rewrite the whole integral in terms of $u$ with no $x$ remaining. If any $x$ survives, your $u$ was the wrong choice. Integrate, then substitute back. For a definite integral, Topic 6.9 is explicit that you either convert the limits to $u$-values or switch fully back to $x$ before evaluating; you cannot plug $x$-limits into a $u$-expression. ### When it is integration by parts (Topic 6.11, BC only) NOTE (BC only): Integration by parts is Topic 6.11 and appears only on the BC exam. AB students can skip this section. Integration by parts reverses the product rule. Reach for it when the integrand is a product of two functions from different families and neither one is the derivative of the other's inside, so substitution has nothing to cancel. That absence of a derivative pairing is the tell. The formula is: $$\int u\,dv = uv - \int v\,du$$ The whole game is choosing $u$ and $dv$ well, and the mnemonic LIATE ranks the choice for $u$: Logarithmic, Inverse trig, Algebraic (polynomials), Trigonometric, Exponential. Whatever appears earliest on that list becomes $u$, the part you differentiate; everything else becomes $dv$, the part you integrate. The point of the ranking is to make $u$ get simpler when differentiated and to leave a $dv$ you can actually integrate. - $\int x e^x\,dx$: polynomial times exponential, so let $u = x$. - $\int x \cos x\,dx$: polynomial times trig, so let $u = x$. - $\int \ln x\,dx$: a lone logarithm, so let $u = \ln x$ and $dv = dx$. - $\int x^2 e^x\,dx$: expect to apply parts twice, since each pass lowers the polynomial's degree by one. ### When it is partial fractions (Topic 6.12, BC only) NOTE (BC only): Linear partial fractions is Topic 6.12 and appears only on the BC exam. The course covers distinct linear, nonrepeating factors, so you will not see repeated factors or irreducible-quadratic decompositions. Partial fractions breaks one complicated rational function into a sum of simple ones you already know how to integrate. Use it when the integrand is a rational function, the denominator factors into distinct linear factors, and the numerator's degree is strictly below the denominator's. Each simple piece integrates to a natural logarithm, which is why the whole technique reduces a hard-looking fraction to a sum of logs. The method: factor the denominator, then write the fraction as a sum of unknown constants over each linear factor, for example $\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$. Clear the denominators, then solve for the constants. Substituting the roots one at a time is the fastest way, because each root kills every term but one and hands you a constant directly. Once you have the constants, you are just integrating a short list of $\frac{1}{x-r}$ terms, each of which becomes $\ln|x-r|$. NOTE (Two checks before you decompose): If the numerator degree is at least the denominator degree, do long division first (Topic 6.10). If the numerator is already the derivative of the denominator, the problem is a plain u-substitution, not partial fractions. ### Common traps and how to avoid them Most lost points on selection problems come from misreading the structure, not from bad algebra. Watch for these. - Skipping the u-substitution check. Integrals that look like parts or partial fractions often collapse under a substitution. $\int \frac{x}{x^2+4}\,dx$ looks like a partial-fractions candidate but is really u-substitution, because the numerator is a constant multiple of the denominator's derivative. - Forgetting to change the limits (or to switch back to $x$) on a definite u-substitution. Topic 6.9 requires the limits to match the variable you are integrating in. - Choosing $u$ and $dv$ backwards in integration by parts, which makes $\int v\,du$ harder than the original integral. Let LIATE decide. - Trying partial fractions when the numerator degree is too high. Divide first, then decompose the remainder. - Dropping the constant of integration $+C$ on an indefinite integral. ### Worked examples #### Recognizing u-substitution in disguise Problem: Evaluate $\int \frac{x}{x^2+4}\,dx$. 1. Classify first. The integrand is a rational function, so partial fractions is tempting. But check the u-substitution signal: is the numerator a multiple of the derivative of the denominator? The denominator is $x^2+4$, and $\frac{d}{dx}(x^2+4) = 2x$. The numerator $x$ is exactly half of that, so this is u-substitution, not partial fractions. 2. Set $u = x^2+4$. Then $du = 2x\,dx$, so $x\,dx = \frac{1}{2}\,du$. 3. Rewrite the integral entirely in $u$, leaving no $x$ behind: $\int \frac{x}{x^2+4}\,dx = \int \frac{1}{u}\cdot \frac{1}{2}\,du = \frac{1}{2}\int \frac{1}{u}\,du$. 4. Integrate the basic form: $\frac{1}{2}\int \frac{1}{u}\,du = \frac{1}{2}\ln|u| + C$. 5. Substitute back. Since $x^2+4 > 0$ for every $x$, the absolute value is optional: $\frac{1}{2}\ln(x^2+4) + C$. Answer: $\frac{1}{2}\ln(x^2+4) + C$ #### Integration by parts: polynomial times exponential (BC) Problem: Evaluate $\int x e^{2x}\,dx$. 1. Classify. The integrand is a product of an algebraic factor $x$ and an exponential $e^{2x}$. Neither is the derivative of the other's inside, so a substitution has nothing to cancel. That is the integration-by-parts signal (BC only, Topic 6.11). 2. Apply LIATE. Algebraic outranks Exponential, so let $u = x$ (the part you differentiate) and $dv = e^{2x}\,dx$ (the part you integrate). 3. Compute the pieces: $du = dx$, and $v = \int e^{2x}\,dx = \frac{1}{2}e^{2x}$. 4. Substitute into $\int u\,dv = uv - \int v\,du$: $\int x e^{2x}\,dx = x\cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x}\,dx$. 5. The remaining integral is basic: $\int \frac{1}{2}e^{2x}\,dx = \frac{1}{2}\cdot \frac{1}{2}e^{2x} = \frac{1}{4}e^{2x}$. 6. Combine and add the constant: $\frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C$. Answer: $\frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C$ #### Partial fractions with distinct linear factors (BC) Problem: Evaluate $\int \frac{1}{x^2+x-2}\,dx$. 1. Classify. This is a rational function with a constant numerator, so the numerator degree (0) is below the denominator degree (2). Factor the denominator: $x^2+x-2 = (x+2)(x-1)$, two distinct linear factors. The numerator is not the derivative of the denominator, since $\frac{d}{dx}(x^2+x-2) = 2x+1$ is not a constant, so this is partial fractions (BC only, Topic 6.12), not u-substitution. 2. Set up the decomposition: $\frac{1}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}$. 3. Clear denominators by multiplying both sides by $(x+2)(x-1)$: $1 = A(x-1) + B(x+2)$. 4. Solve for the constants using the roots. Let $x = 1$: $1 = A(0) + B(3)$, so $B = \frac{1}{3}$. Let $x = -2$: $1 = A(-3) + B(0)$, so $A = -\frac{1}{3}$. 5. Rewrite and integrate each piece: $\int \left(\frac{-1/3}{x+2} + \frac{1/3}{x-1}\right)\,dx = -\frac{1}{3}\ln|x+2| + \frac{1}{3}\ln|x-1| + C$. 6. Combine the logs into one term: $\frac{1}{3}\ln\left|\frac{x-1}{x+2}\right| + C$. Answer: $\frac{1}{3}\ln\left|\frac{x-1}{x+2}\right| + C$ ### FAQ Q: Is integration by parts on the AP Calculus AB exam? A: No. Integration by parts (Topic 6.11) and partial fractions (Topic 6.12) are BC only. AB students handle antidifferentiation with u-substitution (Topic 6.9), the basic antiderivative rules, long division, and completing the square (Topic 6.10). Q: How do I know whether to use u-substitution or integration by parts? A: Check for a derivative pairing first. If one factor is a constant multiple of the derivative of an inside function, use u-substitution. If the integrand is a product of two unrelated functions with no such pairing, use integration by parts. When in doubt, always try the substitution before parts. Q: What is LIATE and when do I use it? A: LIATE stands for Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. In integration by parts, it tells you which factor to call u: pick the one earliest on the list to differentiate, and let the rest be dv. It only matters on BC, since parts is a BC topic. Q: Can you use partial fractions when the denominator does not factor into linear pieces? A: Not on the AP exam. Topic 6.12 covers only distinct linear, nonrepeating factors. An irreducible quadratic denominator is handled by completing the square (Topic 6.10), which usually leads to an arctan rather than a log. ## Related Rates: The 5-Step Method for Any Word Problem URL: https://www.calclearn.org/guides/related-rates Last updated: 2026-07-22 Answer first: Solve any related rates problem in five steps: draw and label a diagram, write an equation relating the variables, differentiate both sides with respect to time, substitute the given values, and solve for the unknown rate. Substitute numbers only after differentiating, never before. ### How to Spot a Related Rates Problem Related rates problems live in CED Topics 4.4 and 4.5, and they all share a signature: two or more quantities change together over time, you are told how fast one of them changes, and you are asked how fast another changes. Watch for the phrase how fast, for a rate given in units per second or per minute, and for a moving picture (a ladder sliding, a balloon inflating, water draining). If you can describe the scene as a short film rather than a frozen snapshot, it is almost certainly related rates. The method-selection question students miss is telling related rates apart from optimization. Both are word problems dressed in geometry, but optimization asks for a largest or smallest value and sets a derivative equal to zero. Related rates asks for a rate at a single instant and never sets a derivative to zero. If the question ends in how fast or at what rate, you are finding the value of a derivative, not hunting for where it vanishes. The engine underneath every related rates problem is the chain rule (CED CHA-3.D.1). Each quantity in the picture is secretly a function of time, so when you differentiate a geometric relationship with respect to $t$, the chain rule attaches a rate factor like $\frac{dx}{dt}$ or $\frac{dh}{dt}$ to every variable. That single idea is why the whole technique works, and it is why we write everything in Leibniz notation: the notation keeps the with-respect-to-$t$ visible at every step. ### The 5-Step Method Every related rates problem, no matter how the scene is dressed up, yields to the same five steps. Run them in order and you never have to improvise. 1. Draw and label. Sketch the situation and assign a letter to every quantity that changes. Immediately write down the rate you know and the rate you want in derivative form, for example $\frac{dx}{dt} = 2$ and find $\frac{dy}{dt}$. 2. Find a governing equation. Write one equation that ties your variables together: a geometry formula, the Pythagorean theorem, similar triangles, or a trig ratio. This is the step that rewards knowing your area and volume formulas cold. 3. Differentiate both sides with respect to $t$. Treat every variable as a function of time, so the chain rule attaches a rate factor to each. This is implicit differentiation applied to time. 4. Substitute the known values. Only now do you plug in the instantaneous numbers, the position or length or radius at the moment in question, along with the given rate. 5. Solve and interpret. Isolate the unknown rate, attach correct units, and read the sign. A negative rate means the quantity is shrinking. ### Substitute Only After You Differentiate NOTE (The one rule that saves the most points): Differentiate first, substitute second. If you plug the instant's numbers into the equation before you differentiate, you freeze a moving variable into a constant, and its derivative collapses to zero. The result is a wrong answer that looks clean, which makes it hard to catch on a timed exam. Here is the trap in slow motion. In the ladder problem below, the base sits $5$ ft from the wall at the instant we care about. If you write $x = 5$ in the equation $x^2 + y^2 = 169$ first, then $x$ is no longer a variable, $\frac{dx}{dt}$ disappears, and you have thrown away the very rate you were given. Differentiate while $x$ is still a live variable, and only then set $x = 5$. There is one honest exception, and it trips people the other way. A relationship that holds at every instant, not just the one in question, can be substituted before differentiating. In the cone problem, similar triangles give $r = \frac{h}{2}$ at all times, so replacing $r$ before differentiating is not only legal, it is the standard move that reduces two variables to one. The test is simple: if a value is true forever (a fixed ratio, a constant length), substitute it early; if it is true only at this instant (say $h = 4$ right now), it must wait until after you differentiate. Three smaller mistakes cost the rest of the points. First, forgetting a chain rule factor when you differentiate a squared or cubed variable. Second, mixing units, such as leaving a radius in centimeters while a rate is in meters. Third, answering the wrong question: the problem may ask for the rate the area grows while you solved for the rate the radius grows. Re-read the final sentence of the prompt before you box your answer. ### Worked examples #### Ladder Sliding Down a Wall Problem: A 13 ft ladder leans against a vertical wall. The base is pulled away from the wall at a constant 2 ft/s. How fast is the top of the ladder sliding down the wall at the moment the base is 5 ft from the wall? 1. Draw and label. Let $x$ be the distance from the wall to the base and $y$ the height of the top on the wall. The ladder length is fixed at $13$ ft. We are given $\frac{dx}{dt} = 2$ ft/s and want $\frac{dy}{dt}$ when $x = 5$. 2. Find a governing equation. The wall, the ground, and the ladder form a right triangle, so the Pythagorean theorem relates the variables: $x^2 + y^2 = 13^2 = 169$. 3. Differentiate both sides with respect to $t$. Every variable carries a chain rule factor: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$. Dividing by $2$ gives $x\frac{dx}{dt} + y\frac{dy}{dt} = 0$. 4. Find the missing length, then substitute. When $x = 5$, the Pythagorean relation gives $y = \sqrt{169 - 25} = \sqrt{144} = 12$. Now substitute $x = 5$, $y = 12$, and $\frac{dx}{dt} = 2$: $(5)(2) + (12)\frac{dy}{dt} = 0$. 5. Solve and interpret. From $10 + 12\frac{dy}{dt} = 0$ we get $\frac{dy}{dt} = -\frac{10}{12} = -\frac{5}{6}$ ft/s. The negative sign confirms the height is decreasing, so the top slides down at $\frac{5}{6}$ ft/s, about $0.83$ ft/s. Answer: The top of the ladder slides down at 5/6 ft/s (about 0.83 ft/s); the negative sign shows the height is decreasing. #### Water Filling a Conical Tank Problem: Water flows into an inverted cone (vertex down) at 3 cubic feet per minute. The tank is 12 ft tall with a radius of 6 ft at the top. How fast is the water level rising when the water is 4 ft deep? 1. Draw and label. Let $V$ be the volume of water, $h$ its depth, and $r$ the radius of the water's surface. We are given $\frac{dV}{dt} = 3$ ft³/min and want $\frac{dh}{dt}$ when $h = 4$. 2. Find a governing equation. The water itself forms a cone, so its volume is $V = \frac{1}{3}\pi r^2 h$. 3. Reduce to one variable using a constant relationship. The water cone is similar to the tank, so $\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$, giving $r = \frac{h}{2}$ at every instant. Because this holds at all times, substitute it now, before differentiating: $V = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi\cdot\frac{h^2}{4}\cdot h = \frac{\pi}{12}h^3$. 4. Differentiate both sides with respect to $t$: $\frac{dV}{dt} = \frac{\pi}{12}\cdot 3h^2\frac{dh}{dt} = \frac{\pi}{4}h^2\frac{dh}{dt}$. 5. Substitute the instantaneous values and solve. Now, and only now, set $h = 4$ and $\frac{dV}{dt} = 3$: $3 = \frac{\pi}{4}(4)^2\frac{dh}{dt} = 4\pi\frac{dh}{dt}$. Therefore $\frac{dh}{dt} = \frac{3}{4\pi}$ ft/min, roughly $0.24$ ft/min. Answer: The water level rises at 3/(4π) ft/min, about 0.24 ft/min. ### FAQ Q: Why do you substitute values only after differentiating? A: Because differentiation acts on variables, not constants. If you replace a changing quantity with its instantaneous value before differentiating, its derivative becomes zero and you lose the rate that quantity contributes. Differentiate while everything is still a variable, then substitute the instant's numbers. The one exception is a relationship true at all times, such as $r = \frac{h}{2}$ from similar triangles, which you may substitute early to reduce variables. Q: Is related rates on the AB or BC exam? A: Both. Related rates is CED Unit 4 (Topics 4.4 and 4.5), which is shared AB and BC content with no BC-only material. Unit 4 makes up 10-15% of the AB multiple-choice section and 5-10% of the BC multiple-choice section. Related rates also appears on free-response questions, so prepare for both formats. Q: What is the difference between related rates and optimization? A: Related rates finds how fast a quantity changes at a specific instant by differentiating with respect to time. Optimization finds a maximum or minimum by setting a derivative equal to zero. If the question asks how fast or at what rate, it is related rates. If it asks for the largest or smallest value, it is optimization. ## Optimization Word Problems: Setup, Constraint, Solve URL: https://www.calclearn.org/guides/optimization-problems Last updated: 2026-07-22 Answer first: Solve an optimization problem by using the constraint (the fixed limitation) to rewrite the objective (the quantity you are maximizing or minimizing) as a function of one variable, setting its derivative equal to zero, and justifying the extreme value with a derivative test. ### Spotting an optimization problem Optimization problems ask for the single best value of something: the largest area, the smallest cost, the most volume, the shortest distance, the least material. That superlative is the tell. Topics 5.10 and 5.11 build directly on the extrema work from Topics 5.2 through 5.7, so once the function is written down, the calculus is the max/min routine you already know. The hard part, and the part almost every missed problem comes down to, is turning the sentence into that function. Two quantities are always in play: one varies and one is pinned down. The quantity that varies and that you want at its extreme is the objective. The quantity that is fixed (a set amount of fencing, a required volume, a sheet of a given size) is the constraint. Read the problem once just to name these two before you write any math. ### Objective vs constraint: the setup that trips everyone The objective function is the thing the question asks you to make as big or as small as possible. If the sentence ends in maximize the area, the objective is area; if it ends in minimize the cost, the objective is cost. Write it as a formula: $A = xy$, $V = \ell w h$, $S = x^2 + 4xh$. It usually starts with two variables, and that is fine for now. The constraint is the fixed number the problem hands you. The sentence 'you have 400 feet of fencing' is a constraint for a three-sided pen (one side runs along a river and needs no fence): $2x + y = 400$. The sentence 'the box must hold 32,000 cubic centimeters' is a constraint: $x^2 h = 32000$. The constraint is almost always an equation, and its job is to connect the two variables that appear in the objective. $$\underbrace{A = xy}_{\text{objective}} \qquad \underbrace{2x + y = 400}_{\text{constraint}}$$ Here is the move that solves every one of these problems: solve the constraint for one variable and substitute it into the objective. That collapses the objective to a function of a single variable, which is exactly what a derivative test needs. If you ever find yourself trying to differentiate a formula that still has two variables in it, you skipped this step. Reading the constraint out of a fence problem versus a box problem works the same way. In a fence problem the fixed total length of fencing is the constraint and the enclosed area is the objective. In a box built from a flat sheet, the sheet size is baked into the dimensions themselves, so there may be no separate constraint equation to write; in a box of required volume, that volume is the constraint and the surface area (material) is the objective. NOTE (Quick test: which is which?): You optimize the objective; you never optimize the constraint. The objective is what the question asks for (largest, smallest, most, least). The constraint is a fixed fact about the situation, stated as a given number. ### The method, step by step Every optimization problem follows the same five moves. Do them in order and the setup stops being the scary part. 1. Draw and label. Assign a variable to every unknown length and sketch the figure. A picture prevents almost every setup error. 2. Write the objective function: the quantity to be maximized or minimized, as a formula. 3. Write the constraint equation, solve it for one variable, and substitute so the objective depends on one variable only. State the domain (the values that make physical sense). 4. Differentiate and find the critical points: set the derivative equal to zero and solve, and note anywhere it fails to exist. 5. Justify that the critical point gives the extreme value you want, then answer the actual question with units. Do not skip the domain in step three. Physical quantities cannot be negative, and a length like $x$ in a $12$ by $12$ sheet cannot exceed $6$ once you cut squares from both ends. The domain tells you which critical points to throw out and, when it is a closed interval, which endpoints you must check. ### Justifying the max or min (AP requires it) Finding the critical point is not the answer. The exam awards a specific point for showing that the critical point is actually the maximum or minimum you claim, and a numeric answer with no justification loses it. You have three legitimate tools. - First derivative test: show $f'(x)$ changes from positive to negative at the critical point (a maximum) or from negative to positive (a minimum). - Second derivative test: show $f''(x)$ is negative at the critical point (concave down, a maximum) or positive (concave up, a minimum). - Candidates test: if the domain is a closed interval, evaluate the objective at every critical point and at both endpoints, then pick the largest or smallest value. There is a shortcut worth knowing from Topic 5.5: if the objective is continuous and has exactly one critical point on the interval, and that point is a local extremum, then it is the global extremum too. That single-critical-point argument is enough justification on its own, and most optimization problems are built to have exactly one critical point in their domain. NOTE (Common mistakes): Differentiating the objective before substituting, so it still has two variables. Forgetting the domain, then reporting a critical point that is negative or otherwise impossible. Answering the wrong quantity: the problem asks for the maximum area and you stop at the value of $x$. Skipping the justification. On free-response that is a lost point every single time. ### Worked examples #### Fence against a river: maximize area Problem: A farmer has 400 feet of fencing to build a rectangular pen along a straight river. The side along the river needs no fence. What dimensions give the largest area? 1. Let $x$ be the length of each side perpendicular to the river and $y$ the side parallel to it. Only three sides are fenced, so the fencing used is $2x + y$. 2. Constraint (fixed fencing): $2x + y = 400$. Solve for $y$: $y = 400 - 2x$. 3. Objective (area to maximize): $A = xy$. Substitute the constraint: $A(x) = x(400 - 2x) = 400x - 2x^2$. 4. Domain: $y > 0$ forces $400 - 2x > 0$, so $0 < x < 200$. 5. Differentiate and find the critical point: $A'(x) = 400 - 4x$. Set $A'(x) = 0$, so $400 - 4x = 0$ and $x = 100$. 6. Justify the maximum with the second derivative test: $A''(x) = -4 < 0$, so the graph is concave down and $x = 100$ is a maximum. 7. Answer the question: $y = 400 - 2(100) = 200$ and $A = (100)(200) = 20{,}000$. The pen is 100 ft by 200 ft. Answer: 100 ft by 200 ft, giving a maximum area of 20,000 square feet. #### Open-top box from a sheet: maximize volume Problem: An open-top box is made from a 12-inch by 12-inch sheet of cardboard by cutting equal squares of side $x$ from each corner and folding up the sides. What value of $x$ maximizes the volume? 1. Let $x$ be the side of each square cut from the corners. After folding, the base is $(12 - 2x)$ by $(12 - 2x)$ and the height is $x$. 2. Objective (volume to maximize): $V = x(12 - 2x)^2$. There is no separate constraint equation here: the sheet size is already built into the dimensions. 3. Domain: the base side must be positive, so $12 - 2x > 0$, giving $0 < x < 6$. 4. Expand: $(12 - 2x)^2 = 144 - 48x + 4x^2$, so $V(x) = 144x - 48x^2 + 4x^3$. 5. Differentiate: $V'(x) = 144 - 96x + 12x^2 = 12(x^2 - 8x + 12) = 12(x - 2)(x - 6)$. 6. Critical points: $x = 2$ or $x = 6$. Since $x = 6$ is outside the domain (it gives zero volume), the only candidate is $x = 2$. 7. Justify with the first derivative test: for $x < 2$, $(x-2)(x-6) > 0$ so $V'(x) > 0$; for $2 < x < 6$, $(x-2)(x-6) < 0$ so $V'(x) < 0$. $V'(x)$ changes from positive to negative at $x = 2$, a maximum. 8. Answer: $V(2) = 2(12 - 4)^2 = 2(8)^2 = 128$. Cut 2-inch squares. Answer: Cut squares of side 2 inches, giving a maximum volume of 128 cubic inches. #### Box of fixed volume: minimize material Problem: A box with a square base and an open top must hold 32,000 cubic centimeters. Find the dimensions that use the least material. 1. Let $x$ be the side of the square base and $h$ the height. The volume is $x^2 h$, and with a base plus four sides (open top) the surface area is $x^2 + 4xh$. 2. Constraint (fixed volume): $x^2 h = 32000$. Solve for $h$: $h = \dfrac{32000}{x^2}$. 3. Objective (material to minimize): $S = x^2 + 4xh$. Substitute: $S(x) = x^2 + 4x \cdot \dfrac{32000}{x^2} = x^2 + \dfrac{128000}{x}$. 4. Domain: $x > 0$. 5. Differentiate: $S'(x) = 2x - \dfrac{128000}{x^2}$. Set $S'(x) = 0$: $2x = \dfrac{128000}{x^2}$. 6. Solve: $2x^3 = 128000$, so $x^3 = 64000$ and $x = 40$. 7. Justify the minimum: $S''(x) = 2 + \dfrac{256000}{x^3} > 0$ for all $x > 0$, so the graph is concave up and the single critical point $x = 40$ is the absolute minimum. 8. Answer: $h = \dfrac{32000}{40^2} = \dfrac{32000}{1600} = 20$ and $S = 40^2 + \dfrac{128000}{40} = 1600 + 3200 = 4800$. Answer: A 40 cm by 40 cm base with height 20 cm, using 4,800 square centimeters of material. ### FAQ Q: How do I tell the objective from the constraint? A: The objective is the quantity the problem asks you to maximize or minimize; the constraint is the fixed amount you are given. You optimize the objective and use the constraint to eliminate a variable. If a sentence says find the largest volume, volume is the objective; if it says using 60 feet of wire, that 60 is the constraint. Q: Do I have to justify the maximum or minimum on the AP exam? A: Yes. On free-response, a correct answer with no justification does not earn full credit. Show a first derivative test, a second derivative test, or a candidates test, or argue that a single critical point on the interval must be the global extremum. Q: What if the problem has no closed interval? A: Use the first or second derivative test, or the single-critical-point argument. The candidates test needs endpoints; when the domain is open (like $x > 0$), rely on the sign of $f'(x)$ or $f''(x)$ instead. Q: Is optimization on both AP Calculus AB and BC? A: Yes. Topics 5.10 and 5.11 are part of Unit 5, which both AB and BC cover, so optimization can appear on either exam. ## Which Series Convergence Test Should I Use? URL: https://www.calclearn.org/guides/which-convergence-test Last updated: 2026-07-22 Answer first: Run a fixed checklist, starting with the nth-term test: if the terms do not shrink to zero, the series diverges. Otherwise match the shape: a geometric ratio, a p-series, alternating signs (alternating series test), factorials or exponentials (ratio test), or a rational function of n (limit or direct comparison). ### Start with the shape, not the test The skill AP Calculus BC actually rewards is recognition: reading the terms of $\sum a_n$ and knowing which test the College Board expects. Series live in Unit 10, worth 15 to 20 percent of the BC exam, and the convergence tests sit in Topics 10.1 through 10.9. There is no single test that works on everything, so trained students run a fixed checklist instead of guessing. Work the list top to bottom and stop at the first test whose conditions the series clearly meets. The order is deliberate: the cheapest checks (a limit, a ratio you can eyeball) come first, and the tests that need setup (comparison, integral) come last. 1. nth-term test first. Compute $\lim_{n \to \infty} a_n$. If it is not zero (or does not exist), the series diverges and you are finished. If the limit is zero, the test is inconclusive, so keep reading. This is Topic 10.3. 2. Geometric? If each term is a fixed multiple of the one before it, so $a_n = a r^n$, the series diverges when $|r| \ge 1$ and converges when $|r| < 1$. The closed form $\sum_{n=0}^{\infty} a r^n = \frac{a}{1-r}$ uses the first term as the numerator, so a sum that starts at $n = 1$ converges to $\frac{ar}{1-r}$ (Topic 10.2). 3. p-series? If the terms look like $\frac{1}{n^p}$, the series converges exactly when $p > 1$ and diverges when $p \le 1$. The harmonic series is the $p = 1$ case, and it diverges (Topic 10.5). 4. Alternating? If the signs flip through $(-1)^n$ or $(-1)^{n-1}$, try the alternating series test, then decide whether the convergence is absolute or conditional (Topics 10.7 and 10.9). 5. Factorials, or $n$ sitting in an exponent? Reach for the ratio test (Topic 10.8). It is built for terms that multiply or divide by a growing factor each step. 6. A messy rational or algebraic function of $n$? Match it to a p-series using the direct comparison or limit comparison test (Topic 10.6). 7. Do the terms come from a positive, decreasing function you can integrate? Use the integral test (Topic 10.4). NOTE (The one-way trap): The nth-term test can only prove divergence. If $\lim_{n \to \infty} a_n = 0$, you have learned nothing yet: the harmonic series has terms going to zero and still diverges. Never write "the terms go to zero, so it converges." That sentence loses points. ### The eight tests at a glance This table is the checklist compressed into a reference. The middle column is the recognition cue (what makes you pick that test), and the last column is the exact condition plus the verdict. These eight are the only convergence tools assessed on the BC exam. The CED covers the six named tests (nth-term, integral, comparison, limit comparison, alternating series, ratio) across Topics 10.3 through 10.9, and the geometric-series (Topic 10.2) and p-series (Topic 10.5) facts round out the eight tools you actually use. | Test | Reach for it when | Condition and verdict | | --- | --- | --- | | nth-term test (10.3) | Always, as the first check | If $\lim_{n \to \infty} a_n \ne 0$, diverges. If the limit is $0$, inconclusive. | | geometric series (10.2) | Constant ratio $r$ between terms | Converges to $\frac{a}{1-r}$ (first term $a$, sum from $n = 0$) if $|r| < 1$; diverges if $|r| \ge 1$. | | p-series (10.5) | Terms are $\frac{1}{n^p}$ | Converges if $p > 1$; diverges if $p \le 1$ (harmonic is $p = 1$). | | alternating series (10.7) | Signs alternate: $\sum (-1)^n b_n$, $b_n > 0$ | Converges if $b_n$ decreases and $\lim_{n \to \infty} b_n = 0$. | | ratio test (10.8) | Factorials, exponentials, $n$th powers | $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$: $L < 1$ converges, $L > 1$ diverges, $L = 1$ inconclusive. | | comparison test (10.6) | Terms bounded above or below by a known series | If $0 \le a_n \le b_n$ and $\sum b_n$ converges, so does $\sum a_n$; if $\sum a_n$ diverges, so does $\sum b_n$. | | limit comparison test (10.6) | Rational or algebraic in $n$, close to a p-series | $L = \lim_{n \to \infty} \frac{a_n}{b_n}$ with $0 < L < \infty$: both series do the same thing. | | integral test (10.4) | $a_n = f(n)$, with $f$ positive and decreasing | $\sum a_n$ and $\int_k^\infty f(x)\,dx$ converge or diverge together. | ### Reading the terms: recognition triggers Most lost time on series problems comes from starting the wrong test, so the cue matters more than the mechanics. Here is what each pattern in the terms is telling you. Factorials or exponentials point to the ratio test. When $a_n$ contains $n!$, $2^n$, or $n^n$, forming $\frac{a_{n+1}}{a_n}$ cancels most of the mess and leaves a clean limit. Factorials almost always win the growth race, so $\sum \frac{3^n}{n!}$ converges while $\sum \frac{n!}{3^n}$ diverges. A rational function of $n$ points to comparison. When $a_n$ is a ratio of polynomials, its long-run size is set by the leading powers. Strip everything but the highest power on top and bottom to get the p-series it behaves like, then confirm with limit comparison. For example $\frac{2n+1}{n^3+5}$ behaves like $\frac{2n}{n^3} = \frac{2}{n^2}$, a convergent p-series. A constant multiplier between terms points to geometric. If dividing any term by the previous one gives the same number every time, it is geometric and you can read off convergence from $|r|$ instantly, no other test needed. Alternating signs let you use the gentlest test. The alternating series test only asks two things, that $b_n$ decreases and that $b_n \to 0$, so a series that fails the ratio or comparison tests may still converge as an alternating series. That is exactly how conditional convergence happens. Terms that match a clean function point to the integral test. If $a_n = f(n)$ for an $f$ you can actually antidifferentiate, and $f$ is positive and decreasing, the improper integral $\int_k^\infty f(x)\,dx$ settles the series. It is the natural choice for $\sum \frac{1}{n \ln n}$ and similar terms with a logarithm. ### Alternating series: absolute versus conditional Once a series converges by the alternating series test, Topic 10.9 asks a follow-up: is the convergence absolute or conditional? Test the series of absolute values $\sum |a_n|$. - If $\sum |a_n|$ converges, the original series is absolutely convergent. Absolute convergence is the stronger statement: if a series converges absolutely, it converges. - If $\sum |a_n|$ diverges but $\sum a_n$ still converges (by the alternating series test), the series is conditionally convergent. The alternating harmonic series $\sum \frac{(-1)^{n-1}}{n}$ is the textbook conditional case: it converges, but stripping the signs gives the harmonic series, which diverges. A useful shortcut on the exam is to run the ratio test on $|a_n|$ first. If the ratio test returns $L < 1$, you already have absolute convergence and can skip the alternating series test entirely. ### Where students lose points - Using the nth-term test to claim convergence. It never proves convergence, only divergence. Terms going to zero is necessary, not sufficient. - Stopping when the ratio test gives $L = 1$. That outcome is inconclusive, and it always happens on p-series and rational terms, which is your signal to switch to comparison instead. - Getting the comparison direction backward. To prove convergence you need a larger convergent series above your terms; to prove divergence you need a smaller divergent series below them. Bounding a series above by a divergent one proves nothing. - Forgetting to check that $b_n$ decreases in the alternating series test. Both conditions are required, and the free-response rubric expects you to state that $b_n$ is decreasing and that $b_n \to 0$. - Misremembering the p-series cutoff. Convergence needs $p > 1$ strictly; $p = 1$ is the harmonic series and diverges. On the exam $p = \frac{1}{2}$ (a common $\frac{1}{\sqrt{n}}$ term) diverges. ### Worked examples #### nth-term test settles it in one line Problem: Determine whether $\sum_{n=1}^{\infty} \frac{3n^2}{5n^2 + 2}$ converges or diverges. 1. The terms are a ratio of polynomials of equal degree, so the very first checklist item, the nth-term test, is worth trying: does $a_n \to 0$? 2. Divide numerator and denominator by $n^2$ to rewrite the terms: $\lim_{n \to \infty} \frac{3n^2}{5n^2 + 2} = \lim_{n \to \infty} \frac{3}{5 + \frac{2}{n^2}}$. 3. As $n \to \infty$ the $\frac{2}{n^2}$ term drops out, so the limit is $\frac{3}{5}$. 4. The limit is $\frac{3}{5}$, which is not zero. By the nth-term test for divergence (Topic 10.3), the series diverges. No further test is needed. Answer: Diverges by the nth-term test, since $\lim_{n \to \infty} a_n = \frac{3}{5} \ne 0$. #### Limit comparison for a rational term Problem: Determine whether $\sum_{n=1}^{\infty} \frac{2n+1}{n^3 + 5}$ converges or diverges. 1. First run the nth-term test: $\lim_{n \to \infty} \frac{2n+1}{n^3+5} = 0$, so that test is inconclusive and we continue down the checklist. 2. The term is rational in $n$, so compare it to a p-series. The leading behavior is $\frac{2n}{n^3} = \frac{2}{n^2}$, which points to $b_n = \frac{1}{n^2}$, a convergent p-series with $p = 2 > 1$. 3. Apply the limit comparison test by forming the ratio: $\frac{a_n}{b_n} = \frac{2n+1}{n^3+5} \cdot \frac{n^2}{1}$. 4. Multiply out the numerator: $\frac{(2n+1)n^2}{n^3+5} = \frac{2n^3 + n^2}{n^3 + 5}$. 5. Take the limit by dividing top and bottom by $n^3$: $L = \lim_{n \to \infty} \frac{2 + \frac{1}{n}}{1 + \frac{5}{n^3}} = \frac{2}{1} = 2$. 6. Since $L = 2$ is finite and positive, $0 < L < \infty$, the two series share a fate. Because $\sum \frac{1}{n^2}$ converges, the original series converges too. Answer: Converges by limit comparison with the convergent p-series $\sum \frac{1}{n^2}$ (limit $L = 2$). #### Ratio test for a factorial term Problem: Determine whether $\sum_{n=1}^{\infty} \frac{3^n}{n!}$ converges or diverges. 1. The presence of a factorial is the recognition cue for the ratio test (Topic 10.8): forming $\frac{a_{n+1}}{a_n}$ will cancel most of the factorial. 2. Write the next term: $a_{n+1} = \frac{3^{n+1}}{(n+1)!}$. Then $\frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{(n+1)!} \cdot \frac{n!}{3^n}$. 3. Rewrite the pieces using $3^{n+1} = 3 \cdot 3^n$ and $(n+1)! = (n+1) \cdot n!$. 4. Substitute and cancel the common $3^n$ and $n!$: $\frac{3 \cdot 3^n}{(n+1) \cdot n!} \cdot \frac{n!}{3^n} = \frac{3}{n+1}$. 5. The terms are positive, so the absolute values change nothing. Take the limit: $L = \lim_{n \to \infty} \frac{3}{n+1} = 0$. 6. Since $L = 0 < 1$, the ratio test gives absolute convergence, and absolute convergence implies convergence. Answer: Converges (absolutely) by the ratio test, since $L = \lim_{n \to \infty} \frac{3}{n+1} = 0 < 1$. ### FAQ Q: Which convergence test should I try first? A: The nth-term test, every time. It is one limit and it can end the problem immediately: if $\lim_{n \to \infty} a_n \ne 0$, the series diverges. If the limit is zero, move on, because the test cannot confirm convergence. Q: Can the nth-term test prove a series converges? A: No. It only proves divergence. Terms shrinking to zero is required for convergence but does not guarantee it, as the diverging harmonic series shows. To prove convergence you need a different test. Q: What do I do when the ratio test gives 1? A: $L = 1$ is inconclusive, so the ratio test tells you nothing and you switch tests. This happens on p-series and rational terms; those are best handled by the comparison or limit comparison test instead. Q: Which convergence tests are on the AP Calculus BC exam? A: The CED (Topics 10.2 through 10.9) lists exactly these: the nth-term test for divergence, and the integral, comparison, limit comparison, alternating series, and ratio tests, plus the geometric-series and p-series facts. Other tests such as the root test are not assessed. ## The Ratio Test: How It Works and When to Use It URL: https://www.calclearn.org/guides/ratio-test Last updated: 2026-07-22 Answer first: Compute L, the limit as n approaches infinity of the absolute value of a term over the one before it. If L < 1 the series converges absolutely; if L > 1 or infinite it diverges; if L = 1 it is inconclusive, so switch tests. Use the ratio test when terms have factorials or a base to the n-th power like r^n. ### The ratio test in one move The ratio test looks at how fast a series' terms shrink by comparing each term to the one right before it. Take the absolute value of the ratio of consecutive terms and send $n$ to infinity to get a single number $L$. $$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$$ That one number decides everything: - $L < 1$: the series converges absolutely. - $L > 1$ (including $L = \infty$): the series diverges. - $L = 1$: the test is inconclusive and tells you nothing, so you must use a different test. This is CED Topic 10.8, BC only. Notice the absolute value: the ratio test always checks for absolute convergence, so it works directly on series with negative or alternating terms (Topic 10.9). If $L < 1$, the series converges no matter what the signs are doing. ### Knowing when to reach for it Method selection matters more than mechanics here. The ratio test is built to cancel two structures: factorials and expressions raised to the $n$-th power. When you see an $n!$ anywhere in the term, the ratio test is almost always the fastest route, because $\frac{(n+1)!}{n!} = n+1$ collapses the factorial to a single factor. The same thing happens with an exponential like $r^n$, since $\frac{r^{n+1}}{r^n} = r$. So the recognition trigger is simple: scan the general term $a_n$. If it contains a factorial, a constant base raised to the $n$ (like $2^n$ or $5^n$), or a product of both, start with the ratio test. A term like $n^n$ also triggers the ratio test, but note its ratio does not collapse to a constant, since $\frac{(n+1)^{n+1}}{n^n} \to \infty$. Just as useful is knowing when it will fail. If the term is purely algebraic, a ratio of polynomials in $n$ such as $\frac{1}{n^2}$ or $\frac{n}{n^3+1}$, the ratio of consecutive terms drifts to $1$ and the test is inconclusive. A p-series behaves the same way. For those, skip the ratio test and go straight to a p-series argument, the comparison or limit comparison test, or the integral test. NOTE (Recognition rule): Factorials or something to the $n$-th power: use the ratio test. Rational functions of $n$ or p-series: expect $L = 1$ and use a different test. ### Running the test step by step Once you have decided the ratio test fits, the procedure is mechanical. 1. Write the general term $a_n$ clearly, then write $a_{n+1}$ by replacing every $n$ with $n+1$. 2. Form the ratio $\frac{a_{n+1}}{a_n}$ and take its absolute value. 3. Simplify by multiplying by the reciprocal of $a_n$: flip the denominator and cancel. Use $(n+1)! = (n+1)\cdot n!$ to reduce factorials and $\frac{r^{n+1}}{r^n} = r$ to reduce powers. 4. Take the limit as $n \to \infty$ to find $L$. 5. Compare $L$ to $1$ and state the conclusion. The single most useful identity is $(n+1)! = (n+1)\cdot n!$. It lets the $n!$ cancel with the $(n+1)!$ and leaves just the factor $n+1$, which is exactly why factorial series come out so clean under this test. ### What to do when L = 1 An inconclusive result is not a divergence result. $L = 1$ means the ratio test cannot see the behavior of the series, not that the series fails to converge. You have to reach for another tool. Two situations produce $L = 1$ in practice. The first is algebraic terms and p-series, where the right follow-up is a p-series argument or the limit comparison test. If the leftover series alternates, the alternating series test usually finishes the job; if it is positive with polynomial terms, comparison or the integral test does. The second is the endpoints of a power series. The ratio test hands you the radius of convergence, but at each endpoint the ratio limit lands on $1$, so you test those two $x$-values separately with whatever ordinary series test fits (Topic 10.13). The worked example on the radius and interval of convergence below shows this in full. ### Common mistakes - Dropping the absolute value. The test is stated with $\left| \cdot \right|$; without it a negative ratio can produce a meaningless limit. - Forgetting to flip. Dividing by $a_n$ means multiplying by its reciprocal, so a common slip is to multiply by $a_n$ instead. - Mishandling factorials. $\frac{(n+1)!}{n!} = n+1$, not $1$ and not another factorial. - Calling $L = 1$ a divergence. It is inconclusive, so switch tests. - Skipping endpoints. After the ratio test gives a radius, a power series is not finished until both endpoints are checked separately. ### Worked examples #### Factorial in the denominator Problem: Determine whether $\sum_{n=1}^{\infty} \frac{2^n}{n!}$ converges or diverges. 1. The term has a factorial, so the ratio test is the natural choice. Set $a_n = \frac{2^n}{n!}$ and $a_{n+1} = \frac{2^{n+1}}{(n+1)!}$. 2. Form the absolute ratio and flip the denominator: $\left| \frac{a_{n+1}}{a_n} \right| = \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n}$. 3. Cancel using $\frac{2^{n+1}}{2^n} = 2$ and $\frac{n!}{(n+1)!} = \frac{1}{n+1}$: the ratio becomes $2 \cdot \frac{1}{n+1} = \frac{2}{n+1}$. 4. Take the limit: $\lim_{n \to \infty} \frac{2}{n+1} = 0$, so $L = 0$. 5. Since $L = 0 < 1$, the series converges absolutely. Answer: The series converges absolutely, since $L = 0 < 1$. #### Factorial beats a big exponential Problem: Determine whether $\sum_{n=1}^{\infty} \frac{n!}{100^n}$ converges or diverges. 1. A factorial signals the ratio test. Let $a_n = \frac{n!}{100^n}$, so $a_{n+1} = \frac{(n+1)!}{100^{n+1}}$. 2. Form the absolute ratio: $\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)!}{100^{n+1}} \cdot \frac{100^n}{n!}$. 3. Cancel with $\frac{(n+1)!}{n!} = n+1$ and $\frac{100^n}{100^{n+1}} = \frac{1}{100}$: the ratio is $\frac{n+1}{100}$. 4. Take the limit: $\lim_{n \to \infty} \frac{n+1}{100} = \infty$, so $L = \infty$. 5. Since $L > 1$, the series diverges. The takeaway: a factorial in the numerator overwhelms even a base as large as $100$, because the ratio grows without bound. Answer: The series diverges, since $L = \infty > 1$. #### Radius and interval of convergence Problem: Find all $x$ for which the power series $\sum_{n=1}^{\infty} \frac{x^n}{n}$ converges. 1. This is Topic 10.13: apply the ratio test to the terms, treating $x$ as fixed. Let $a_n = \frac{x^n}{n}$, so $a_{n+1} = \frac{x^{n+1}}{n+1}$. 2. Form the absolute ratio: $\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^{n+1}}{n+1} \cdot \frac{n}{x^n} \right| = |x| \cdot \frac{n}{n+1}$. 3. Take the limit: $\lim_{n \to \infty} |x| \cdot \frac{n}{n+1} = |x|$, so $L = |x|$. 4. The ratio test guarantees convergence when $L < 1$, that is $|x| < 1$. The radius of convergence is $R = 1$, giving the open interval $-1 < x < 1$. 5. At the endpoints $L = 1$, so the ratio test is inconclusive and each endpoint is tested on its own. At $x = 1$ the series is $\sum \frac{1}{n}$, the harmonic series, which diverges. At $x = -1$ the series is $\sum \frac{(-1)^n}{n}$, the alternating harmonic series, which converges by the alternating series test. 6. Combine the results: the series converges for $-1 \le x < 1$. Answer: Converges on $[-1, 1)$: radius $R = 1$, diverging at $x = 1$ and converging at $x = -1$. ### FAQ Q: What does it mean when the ratio test equals 1? A: It means the test is inconclusive: it gives you no information, and the series could still converge or diverge. Do not read $L = 1$ as divergence. Switch to another test. For rational terms or a p-series use a p-series argument or the limit comparison test; for an alternating leftover use the alternating series test; for the endpoints of a power series test each endpoint on its own. Q: Does the ratio test work on alternating series? A: Yes. The test uses the absolute value $\left| \frac{a_{n+1}}{a_n} \right|$, so signs do not matter while you compute $L$. If $L < 1$, the series converges absolutely, which means it converges regardless of the alternating signs (Topic 10.9). If $L = 1$, fall back to the alternating series test. Q: When should I use the ratio test instead of the comparison test? A: Use the ratio test when the term has factorials or something raised to the $n$-th power, since those cancel cleanly in the ratio. Use a comparison or limit comparison test when the term is a rational function of $n$ (polynomials over polynomials), because there the ratio test just gives $L = 1$ and tells you nothing. Q: Is the root test on the AP Calculus BC exam? A: No. The CED lists the tests assessed on the BC exam (nth term, integral, comparison, limit comparison, alternating series, and ratio), and the root test is not among them. It exists and can handle some series the ratio test cannot, but it is beyond the AP scope, so stick with the ratio test for exam work. ## How to Find Limits: Every Method, in Order URL: https://www.calclearn.org/guides/how-to-find-limits Last updated: 2026-07-22 Answer first: Try direct substitution first: plug in the value x approaches, and a real number is your answer. If you get 0/0, factor, or multiply by a conjugate when a square root appears. Match sin x / x shapes to known trig limits, and use the squeeze theorem for oscillating functions. ### The order to try methods The fastest way to find a limit is to substitute the value $x$ approaches. If the function is defined and continuous there, the number you get is the limit, and you are done (Topic 1.5). Everything else in this guide exists for the cases where substitution fails. So the real skill is not memorizing five techniques: it is reading the form substitution hands back and letting that form pick the technique. After you substitute, one of a few things happens. Match what you see to the row below, then jump to that method. This table is the whole decision procedure, and the rest of the guide just fills in each row. | After substituting you get | What it means | What to do | | --- | --- | --- | | A real number | The function is continuous at that point | Stop. That number is the limit. | | $\frac{0}{0}$ | Indeterminate: a common factor is hiding | Factor, or multiply by a conjugate if a radical is present | | A nonzero number over $0$, such as $\frac{5}{0}$ | Vertical asymptote | The limit is $+\infty$, $-\infty$, or does not exist (Topic 1.14) | | $\frac{0}{0}$ from a $\frac{\sin x}{x}$ shape | A special trig limit | Rewrite to a known trig limit, or use the squeeze theorem (Topic 1.8) | NOTE: $\frac{0}{0}$ is called an indeterminate form for a reason: it is a signal to keep working, not an answer. It never means the limit is $0$, and it does not by itself mean the limit fails to exist. ### Direct substitution: the default Substitution works because of the limit laws and continuity. The limit laws (Topic 1.5) say the limit of a sum, difference, product, quotient, or composition is built from the limits of the pieces, and polynomial, rational, power, exponential, logarithmic, and trigonometric functions are continuous everywhere they are defined (Topic 1.12). Put those together and, for any of those functions, the limit as $x \to c$ equals the value at $c$ whenever $c$ is in the domain (continuity at a point, Topic 1.11). So $\lim_{x \to 4} (x^2 - 3x + 1)$ is just $16 - 12 + 1 = 5$. No manipulation, no special cases. If you can plug in the number and nothing blows up, that is the whole problem. Reach for the harder methods only when substitution produces $\frac{0}{0}$. This is why so many textbook limits look trivial: they are continuity in disguise. The interesting problems are engineered so that substitution breaks, which is the exam's way of testing whether you can recognize and repair the break. When it breaks, it almost always breaks as $\frac{0}{0}$, and that specific failure is what the next two sections handle. ### When substitution gives 0/0: factor or use the conjugate If substituting $c$ makes both the numerator and denominator zero, a common factor of $(x - c)$ is hiding. For polynomials, $(x - c)$ divides both of them directly; when a radical is present, that shared $(x - c)$ factor is exposed only after you multiply by the conjugate. Either way, that shared factor is what makes the fraction look like $\frac{0}{0}$, and your job is to expose it and cancel it (Topic 1.6). Factoring is the move when you have polynomials. Factor the top and bottom, cancel the common $(x - c)$, and substitute into what is left. The cancellation is legal because a limit only cares about $x$ near $c$, not at $c$, so dividing by $(x - c)$ is fine for every $x$ the limit actually looks at. The conjugate is the move when a square root sits inside the $\frac{0}{0}$ expression. Multiply the numerator and denominator by the conjugate of the radical part, which turns a difference like $\sqrt{x+4} - 2$ into a difference of squares. The radical disappears, a factor cancels, and substitution finishes the job. Quick recognition rule: polynomials on both sides means factor; a stubborn square root means conjugate. ### Trig limits and the squeeze theorem Some $\frac{0}{0}$ limits will not factor and have no radical to rationalize, because the trouble is trigonometric. Two results, both provable with the squeeze theorem, cover almost every case on the exam: $$\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0$$ When you see a sine divided by its own argument (or something you can reshape into that), match it to the first result. For example, $\lim_{x \to 0} \frac{\sin 3x}{4x}$ is $\frac{0}{0}$, but rewriting gives $\frac{\sin 3x}{4x} = \frac{3}{4} \cdot \frac{\sin 3x}{3x}$. As $x \to 0$ the argument $3x \to 0$ as well, so $\frac{\sin 3x}{3x} \to 1$ and the limit is $\frac{3}{4}$. The squeeze theorem is the tool of last resort, and the one to reach for when a function oscillates. It says that if $g(x) \le f(x) \le h(x)$ for all $x$ near $c$, and $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$, then $\lim_{x \to c} f(x) = L$ too. Use it when substitution and algebra both fail because a bounded but wildly oscillating piece like $\sin\!\left(\frac{1}{x}\right)$ refuses to settle, yet you can trap the whole expression between two functions that meet at the same value (Topic 1.8). ### Pitfalls and how to choose fast - $\frac{0}{0}$ versus $\frac{c}{0}$: a zero over a zero is indeterminate and calls for algebra; a nonzero number over zero is a vertical asymptote, so the limit is infinite or does not exist. They look similar and lead to opposite conclusions. - Do not stop at $\frac{0}{0}$ and write 'does not exist.' That form means the answer is still reachable, usually with one factoring or conjugate step. - Check one-sided limits when the function is piecewise, or when $x$ approaches the edge of a domain or a vertical asymptote. If the left and right limits disagree, the two-sided limit does not exist (Topic 1.3). - L'Hôpital's Rule (BC only, Topic 4.7) is not needed for anything in Unit 1. Every $\frac{0}{0}$ here yields to factoring, a conjugate, or a trig limit. - If a function oscillates without settling, either squeeze it or conclude the limit does not exist. Oscillation never factors away. Run the same loop every time: substitute, read the form, and let the form choose the method. That single habit, not raw memorization, is what makes limit problems fast. ### Worked examples #### Factoring a 0/0 rational limit (Topic 1.6) Problem: Evaluate $\lim_{x \to 2} \frac{x^2 - 4}{x^2 - x - 2}$. 1. Substitute $x = 2$ first. The numerator is $4 - 4 = 0$ and the denominator is $4 - 2 - 2 = 0$, giving $\frac{0}{0}$. That is indeterminate, so both parts share the factor $(x - 2)$. Factor to find it. 2. Factor the numerator as a difference of squares: $x^2 - 4 = (x - 2)(x + 2)$. 3. Factor the denominator: $x^2 - x - 2 = (x - 2)(x + 1)$. 4. Rewrite the fraction and cancel the common factor: $\frac{(x - 2)(x + 2)}{(x - 2)(x + 1)} = \frac{x + 2}{x + 1}$. Canceling $(x - 2)$ is valid because the limit only uses values of $x$ near $2$, never $x = 2$ itself. 5. Substitute $x = 2$ into the reduced form: $\frac{2 + 2}{2 + 1} = \frac{4}{3}$. Answer: The limit is 4/3. #### Rationalizing with a conjugate (Topic 1.6) Problem: Evaluate $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$. 1. Substitute $x = 0$: the numerator is $\sqrt{4} - 2 = 0$ and the denominator is $0$, giving $\frac{0}{0}$. A square root sits in the indeterminate form, so multiply by the conjugate. 2. Multiply the numerator and denominator by $\sqrt{x + 4} + 2$: $\frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}$. 3. The numerator becomes a difference of squares: $\left(\sqrt{x + 4}\right)^2 - 2^2 = (x + 4) - 4 = x$. 4. The expression is now $\frac{x}{x\left(\sqrt{x + 4} + 2\right)}$. Cancel the common factor $x$: $\frac{1}{\sqrt{x + 4} + 2}$. 5. Substitute $x = 0$: $\frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$. Answer: The limit is 1/4. #### Trapping an oscillating function with the squeeze theorem (Topic 1.8) Problem: Evaluate $\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right)$. 1. Substitution and the limit laws fail here: $\sin\!\left(\frac{1}{x}\right)$ oscillates faster and faster as $x \to 0$ and has no limit, so there is no factor to cancel. Look for bounds instead. 2. Bound the sine. For every $x \neq 0$, $-1 \le \sin\!\left(\frac{1}{x}\right) \le 1$. 3. Multiply the inequality through by $x^2$. Since $x^2 \ge 0$, the inequality directions are preserved: $-x^2 \le x^2 \sin\!\left(\frac{1}{x}\right) \le x^2$. 4. Take the limit of the two outer bounds: $\lim_{x \to 0} \left(-x^2\right) = 0$ and $\lim_{x \to 0} x^2 = 0$. 5. The middle expression is squeezed between two functions that both approach $0$, so by the squeeze theorem $\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0$. Answer: The limit is 0. ### FAQ Q: What does 0/0 mean when finding a limit? A: It is an indeterminate form, which means substitution failed but the limit is still findable. The numerator and denominator share a factor of $(x - c)$, directly for polynomials, or after rationalizing when a radical is present; remove it by factoring or by multiplying by a conjugate, then substitute again. Q: How do I know whether to factor or use the conjugate? A: Look at the expression. If the top and bottom are polynomials, factor and cancel. If a square root is part of the $\frac{0}{0}$ form, multiply by the conjugate of the radical to clear it, then cancel. Q: When should I use the squeeze theorem? A: Use it when substitution and algebra both fail because a bounded piece oscillates and never settles, such as $\sin\!\left(\frac{1}{x}\right)$. Trap the whole expression between two functions that approach the same value, and the limit equals that value (Topic 1.8). Q: Can I use L'Hôpital's Rule instead? A: L'Hôpital's Rule is BC-only (Topic 4.7) and AB students never need it here. For Unit 1 limits you do not need it, and it also will not help with oscillating cases, which require the squeeze theorem rather than differentiation. ## L'Hopital's Rule: When You Can and Cannot Use It URL: https://www.calclearn.org/guides/lhopitals-rule Last updated: 2026-07-22 Answer first: L'Hopital's Rule applies only when direct substitution gives 0/0 or infinity/infinity. Verify the form first: a value like 1/1, or a 5/0 form whose limit is infinite or does not exist, means the rule does not apply and gives a wrong answer. When it applies, differentiate top and bottom separately, then re-evaluate. ### Verify the Form Before You Touch the Rule L'Hopital's Rule is a shortcut for evaluating a limit of a quotient $\frac{f(x)}{g(x)}$, but it comes with one hard precondition: direct substitution has to produce the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Checking that form is not a formality you can skip when you are in a hurry. It is the whole rule of admission. If substitution gives anything else, the rule is not valid, and applying it can convert a correct limit into a wrong number. This is Topic 4.7 on the AP Calculus AB and BC exams. The reason the check matters is that the rule is a theorem with a hypothesis, not a universal identity. It says that when $\frac{f(x)}{g(x)}$ approaches $\frac{0}{0}$ or $\frac{\infty}{\infty}$, the original limit equals $\lim_{x \to a} \frac{f'(x)}{g'(x)}$, provided that second limit exists or is infinite. Strip away the hypothesis and the two limits have no reason to be equal. That is why an unchecked application does not merely risk an error: it computes the limit of a genuinely different expression. NOTE (0/0 is a label, not a value): The College Board is explicit that $\frac{0}{0}$ names an indeterminate form; it is not a fraction you can simplify to $1$ or to $0$. Treating it as a number is the exact error Topic 4.7 warns against. It signals only that more work is needed, not that the answer is decided. ### How to Recognize an Indeterminate Form Recognition is mechanical. Substitute the value the variable approaches into the numerator and the denominator on their own, then read the pair of results. Two outcomes, and only two, let you use L'Hopital's Rule: the numerator and denominator both go to $0$, or they both grow without bound. Every other outcome already tells you the limit directly, so there is nothing left to resolve and no reason to differentiate. | Substitution result | Indeterminate? | L'Hopital's allowed? | | --- | --- | --- | | $\frac{0}{0}$ | Yes | Yes | | $\frac{\infty}{\infty}$ | Yes | Yes | | $\frac{5}{0}$ | No (limit is $\pm\infty$ or does not exist) | No | | $\frac{0}{5}$ | No (limit is $0$) | No | | $\frac{3}{2}$ | No (limit is $\frac{3}{2}$) | No | The bottom three rows are worth staring at, because they are where students lose points. A result like $\frac{5}{0}$ means the quotient blows up: the limit is infinite or does not exist, and that is the final answer. A result like $\frac{0}{5}$ equals $0$, and $\frac{3}{2}$ equals $\frac{3}{2}$. In each case substitution handed you the answer. Differentiating the top and bottom would replace that answer with the limit of a different function entirely. A couple of forms tempt students into L'Hopital's Rule even though they are not quotients yet: $0 \cdot \infty$ and $\infty - \infty$. The AP Course and Exam Description (CED) explicitly excludes these from both the AB and BC exams, so you will not be asked to resolve them. If you ever meet one in other coursework, you first rewrite it as a single fraction; only then can a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form appear. ### Applying L'Hopital's Rule Once the form is confirmed, the mechanics are short. Differentiate the numerator and the denominator separately, keep them as a quotient, and evaluate again. The single most common execution error is reaching for the quotient rule. L'Hopital's Rule does not differentiate $\frac{f(x)}{g(x)}$ as one object; it differentiates $f$ and $g$ independently and then divides the results. $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \quad \text{when } \frac{f(x)}{g(x)} \to \frac{0}{0} \text{ or } \frac{\infty}{\infty}$$ After you form the new quotient $\frac{f'(x)}{g'(x)}$, substitute again. If you now get a determinate value, that value is the limit and you are done. If you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ a second time, the rule still applies, so verify the form and differentiate again. You can repeat this as many times as the form stays indeterminate, but you must re-check after every pass, because the moment the form resolves you must stop. One caution on repetition: L'Hopital's Rule promises the answer only when the new limit $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ actually exists or runs to $\pm\infty$. On the AB and BC exams the problems are built so that a few passes resolve cleanly, so if your quotient keeps returning an indeterminate form and never simplifies, that is a signal to try algebra first, such as factoring or dividing by the dominant term, rather than differentiating forever. ### Common Misuse: When You Cannot Use It Almost every wrong use of L'Hopital's Rule traces back to skipping the form check. Here are the misuses that show up most on quizzes and free-response scoring: - The form is not indeterminate. If substitution gives a defined value, a nonzero-over-zero like $\frac{4}{0}$, or a zero-over-nonzero like $\frac{0}{7}$, the limit is already settled. Differentiating changes the problem and usually produces a wrong answer, as the last worked example shows. - Using the quotient rule by mistake. Differentiate the numerator and denominator on their own. The derivative of the whole quotient is a different expression and gives the wrong limit. - Forgetting to re-verify between passes. Each application is legal only if the current quotient is still $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Applying it once more after the form has resolved reintroduces the same error. - Reaching for it when simpler tools work. A limit like $\frac{3x^2 + 2x}{x^2 - 5}$ as $x \to \infty$ is faster to settle by dividing by the highest power, and rational limits at a finite point often factor and cancel. L'Hopital's Rule is legal on many of these, but it is not always the quickest or safest route. The habit that keeps you safe is boring and reliable: before you differentiate anything, substitute and name the form out loud. If it is $\frac{0}{0}$ or $\frac{\infty}{\infty}$, proceed. If it is anything else, you already have your answer or you need a different technique. That one check is what separates a limit L'Hopital's Rule can finish from one it will quietly ruin, and it costs you only a few seconds under exam pressure. The verification step is the method. ### Worked examples #### A 0/0 form Problem: Evaluate $\lim_{x \to 0} \frac{e^x - 1}{x}$. 1. Check the form by substituting $x = 0$: $\frac{e^{0} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$. The form is indeterminate, so the rule is allowed. 2. Name the pieces $f(x) = e^x - 1$ and $g(x) = x$, then differentiate each separately: $f'(x) = e^x$ and $g'(x) = 1$. 3. Replace the original quotient with $\frac{f'(x)}{g'(x)}$ and evaluate: $\lim_{x \to 0} \frac{e^x}{1} = \frac{e^{0}}{1} = \frac{1}{1} = 1$. Answer: 1 #### An infinity/infinity form needing two passes Problem: Evaluate $\lim_{x \to \infty} \frac{x^2}{e^x}$. 1. Read off the behavior as $x \to \infty$: the numerator $x^2 \to \infty$ and the denominator $e^x \to \infty$, giving $\frac{\infty}{\infty}$. This is indeterminate, so apply the rule. 2. With $f(x) = x^2$ and $g(x) = e^x$, differentiate separately: $f'(x) = 2x$ and $g'(x) = e^x$. The new limit is $\lim_{x \to \infty} \frac{2x}{e^x}$. 3. Re-check the form: as $x \to \infty$, $\frac{2x}{e^x} \to \frac{\infty}{\infty}$, still indeterminate. Apply the rule a second time. Differentiating $2x$ gives $2$ and differentiating $e^x$ gives $e^x$, so the limit becomes $\lim_{x \to \infty} \frac{2}{e^x}$. 4. Evaluate the last form: as $x \to \infty$, $e^x \to \infty$, so $\frac{2}{e^x} \to \frac{2}{\infty} = 0$. This is a determinate value, so stop. Answer: 0 #### A limit that looks like a job for L'Hopital's but is not Problem: Evaluate $\lim_{x \to 0} \frac{\cos x}{x + 1}$, and see why L'Hopital's Rule gives the wrong answer here. 1. Always check the form first. Substitute $x = 0$: $\frac{\cos 0}{0 + 1} = \frac{1}{1} = 1$. The form is $\frac{1}{1}$, a defined value, so the limit is simply $1$. Direct substitution already finished the problem. 2. Because the form is not $\frac{0}{0}$ or $\frac{\infty}{\infty}$, L'Hopital's Rule does not apply. Watch what happens if you use it anyway: differentiating the numerator gives $-\sin x$ and the denominator gives $1$, producing $\lim_{x \to 0} \frac{-\sin x}{1} = \frac{-\sin 0}{1} = 0$. 3. The illegal application returned $0$, but the true limit is $1$. The rule changed the answer because the hypothesis, an indeterminate form, was never satisfied. This is exactly the misuse to guard against. Answer: 1 by direct substitution. L'Hopital's Rule does not apply; misusing it wrongly gives 0. ### FAQ Q: Is L'Hopital's Rule on the AP Calculus exam? A: Yes. It is Topic 4.7 and appears on both AB and BC. The exam only asks you to use it on the indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$. Q: Can you use L'Hopital's Rule for 0 times infinity? A: Not directly. The forms $0 \cdot \infty$ and $\infty - \infty$ are excluded from both the AB and BC exams. In other coursework you would first rewrite the expression as a single fraction so that a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form appears, then apply the rule. Q: Do you use the quotient rule with L'Hopital's Rule? A: No. Differentiate the numerator and the denominator separately, then divide. Using the quotient rule on the whole fraction produces a different expression and the wrong limit. Q: How many times can you apply L'Hopital's Rule? A: As many times as the quotient stays $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Re-check the form after each pass and stop the instant it resolves to a determinate value. ## Implicit Differentiation: Finding dy/dx Step by Step URL: https://www.calclearn.org/guides/implicit-differentiation Last updated: 2026-07-22 Answer first: Implicit differentiation finds dy/dx when an equation mixes x and y and you can't solve for y cleanly. Differentiate both sides with respect to x; each y-term picks up a dy/dx from the chain rule. Then gather the dy/dx terms, factor, and divide. Reach for it when y is trapped, like x^2 + y^2 = 25. ### Spotting when to use implicit differentiation Implicit differentiation is the tool when an equation ties $x$ and $y$ together and you cannot (or would rather not) solve for $y$ first. A circle such as $x^2 + y^2 = 25$ is the classic signal: $y$ appears squared, so isolating it forces a $\pm$ and breaks the curve into two separate halves. Equations like $x^3 + y^3 = 6xy$ or $\sin(xy) = x$ are worse still, because $y$ is tangled past anything clean algebra can undo. These are relations, not functions, and many of them fail the vertical line test outright. The recognition question is short. Can you rewrite the equation as $y = f(x)$ without a fight? If yes, do exactly that and differentiate with the ordinary rules from Units 2 and 3. If $y$ is trapped, mixed into products with $x$, or raised to powers you cannot unwind, switch to implicit differentiation. You never isolate $y$ at all. You differentiate the equation as written and solve for the slope $\frac{dy}{dx}$ directly. | Equation | Can you isolate $y$? | Best method | | --- | --- | --- | | $y = x^2\sin x$ | Already solved | Ordinary rules (product rule) | | $x^2 + y^2 = 25$ | Only with a $\pm$ | Implicit differentiation | | $x^3 + y^3 = 6xy$ | No clean way | Implicit differentiation | | $xy = 4$ | Yes, $y = \frac{4}{x}$ | Either works | NOTE (The one idea behind the method): Every $y$ in the equation stands for an unnamed function of $x$. So every $y$ you differentiate carries a hidden $\frac{dy}{dx}$, supplied by the chain rule. That is the entire basis of Topic 3.2, and it is why the chain rule from Topic 3.1 has to come first. ### The method: differentiate, then solve for dy/dx Differentiate both sides of the equation with respect to $x$. Terms built only from $x$ follow the rules you already know, and a constant differentiates to $0$. The moment you differentiate a term containing $y$, the chain rule fires. The derivative of $y^2$ is not $2y$; it is $2y\frac{dy}{dx}$, because you differentiate the outer power and then multiply by the derivative of the inside function: the inside is $y$, and its derivative is $\frac{dy}{dx}$. The general pattern is worth memorizing: $$\frac{d}{dx}\left(y^n\right) = n y^{n-1}\frac{dy}{dx}$$ 1. Differentiate both sides with respect to $x$, treating $y$ as a function of $x$. 2. Apply the chain rule to every $y$-term, attaching a factor of $\frac{dy}{dx}$; use the product rule wherever $x$ and $y$ are multiplied. 3. Move every term containing $\frac{dy}{dx}$ to one side and everything else to the other. 4. Factor out $\frac{dy}{dx}$ and divide to isolate it. Step four explains why the result almost always contains both $x$ and $y$. That is not a mistake to fix. The slope of an implicit curve genuinely depends on which point you sit at, and naming a point on such a curve takes both coordinates. An answer like $\frac{dy}{dx} = -\frac{x}{y}$ is a finished answer. When a problem asks for the slope at a specific point, wait until after you have solved for $\frac{dy}{dx}$, then substitute both coordinates. Substituting too early, before the derivative is isolated, throws away the $\frac{dy}{dx}$ you are trying to find. ### The mistakes that cost points The single most common error is differentiating a $y$-term as if $y$ were $x$. Writing $\frac{d}{dx}(y^3) = 3y^2$ silently drops the $\frac{dy}{dx}$ and corrupts every step after it. Every $y$ you touch earns its chain-rule factor, without exception. The second trap shows up whenever $x$ and $y$ are multiplied, like the $xy$ in $x^2 + xy + y^2 = 7$. That term needs the product rule, and the $y$-half of the product still needs its own $\frac{dy}{dx}$. The correct expansion is $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$. Skipping the product rule here is one of the most frequent free-response deductions in Unit 3. The third slip is algebraic. After you collect the $\frac{dy}{dx}$ terms, factor $\frac{dy}{dx}$ out before you divide. Trying to divide term by term instead of factoring first is where a correct derivative turns into a wrong final fraction. ### Extending to the second derivative To find $\frac{d^2y}{dx^2}$, differentiate the first derivative again with respect to $x$. Two details do the damage. First, the first derivative usually still contains $y$, so differentiating it calls for the chain rule (and often the quotient rule) a second time. Second, any $\frac{dy}{dx}$ that reappears in this pass should be replaced with the expression you already found, after which you simplify, often using the original equation to clean things up. Topic 3.6 formalizes higher-order derivatives; the circle example below runs the whole process end to end. This is also why a second-derivative answer for a circle can collapse to something as tidy as $-\frac{25}{y^3}$: the original relation absorbs the $x^2 + y^2$ that the algebra produces. ### Worked examples #### The circle: a clean first derivative Problem: Find $\frac{dy}{dx}$ for the circle $x^2 + y^2 = 25$, then find the slope at the point $(3, 4)$. 1. Differentiate both sides with respect to $x$, treating $y$ as an unknown function of $x$: $\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$. 2. The $x^2$ term gives $2x$ by the power rule. The $y^2$ term needs the chain rule, so $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$. The constant $25$ differentiates to $0$, leaving $2x + 2y\frac{dy}{dx} = 0$. 3. Move the $x$-term across and divide by $2y$: $2y\frac{dy}{dx} = -2x$, so $\frac{dy}{dx} = -\frac{x}{y}$. 4. Now substitute the point $(3, 4)$, after the derivative is isolated: $\frac{dy}{dx} = -\frac{3}{4}$. Answer: $\frac{dy}{dx} = -\frac{x}{y}$, which is $-\frac{3}{4}$ at $(3, 4)$. #### A product term: xy needs the product rule Problem: Find $\frac{dy}{dx}$ for $x^2 + xy + y^2 = 7$. 1. Differentiate each term with respect to $x$. The middle term $xy$ is a product of two functions of $x$, so flag it for the product rule: $\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = 0$. 2. The power rule gives $\frac{d}{dx}(x^2) = 2x$. The product rule gives $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$, since $\frac{d}{dx}(x) = 1$. The chain rule gives $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$. 3. Assemble the pieces: $2x + x\frac{dy}{dx} + y + 2y\frac{dy}{dx} = 0$. 4. Gather the $\frac{dy}{dx}$ terms on one side and move the rest to the other: $x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y$. 5. Factor out $\frac{dy}{dx}$, then divide: $\frac{dy}{dx}(x + 2y) = -(2x + y)$, so $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$. Answer: $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$. #### The second derivative of the circle Problem: Find $\frac{d^2y}{dx^2}$ for the circle $x^2 + y^2 = 25$. 1. Start from the first derivative already found for this circle, $\frac{dy}{dx} = -\frac{x}{y}$. 2. Differentiate again with respect to $x$. The right side is a quotient, so use the quotient rule, remembering that $\frac{d}{dx}(y) = \frac{dy}{dx}$: $\frac{d^2y}{dx^2} = -\frac{(1)(y) - x\frac{dy}{dx}}{y^2}$. 3. Substitute the known $\frac{dy}{dx} = -\frac{x}{y}$: $\frac{d^2y}{dx^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2}$. 4. Combine the numerator over $y$: $y + \frac{x^2}{y} = \frac{y^2 + x^2}{y}$, so $\frac{d^2y}{dx^2} = -\frac{x^2 + y^2}{y^3}$. 5. The original equation says $x^2 + y^2 = 25$, so the result simplifies to $\frac{d^2y}{dx^2} = -\frac{25}{y^3}$. Answer: $\frac{d^2y}{dx^2} = -\frac{25}{y^3}$. ### FAQ Q: When should I use implicit differentiation instead of solving for y? A: Use it whenever solving for $y$ is impossible or messy: $y$ appears to a power, sits inside a product with $x$, or is buried in a trig or exponential expression. If you can write $y = f(x)$ cleanly, the ordinary rules are faster. If you cannot, differentiate the equation as it stands. Q: Why does dy/dx appear when I differentiate a y-term? A: Because $y$ is a function of $x$. Differentiating $y^2$ is a chain-rule problem: the outer derivative $2y$ times the derivative of the inside, which is $\frac{dy}{dx}$. That $\frac{dy}{dx}$ is the derivative of the inner function, and it never disappears. Q: Why does my answer have both x and y in it? A: That is normal and correct for implicit curves. The slope changes from point to point, and a point on a relation like $x^2 + y^2 = 25$ needs both an $x$ and a $y$ to pin it down. To get a number, substitute a specific point after solving for $\frac{dy}{dx}$. ## The Fundamental Theorem of Calculus: Part 1 vs Part 2 URL: https://www.calclearn.org/guides/fundamental-theorem-of-calculus Last updated: 2026-07-22 Answer first: Part 2 (evaluation) turns a definite integral into a number: find an antiderivative F of f, then the integral from a to b is F(b) minus F(a). Part 1 (accumulation) turns an integral into a new function: the derivative of g(x) = integral from a to x of f(t) dt is just f(x). ### The two statements, side by side Both halves of the Fundamental Theorem of Calculus (FTC) say the same big thing: differentiation and integration undo each other. They run that relationship in opposite directions, which is why they get used for completely different tasks. The evaluation form (AP CED Topic 6.7, essential knowledge FUN-6.B.3) is the one you use to compute integrals. If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$, then: $$\int_a^b f(x)\,dx = F(b) - F(a)$$ The accumulation form (Topic 6.4, FUN-6.B.2) runs the other way. It differentiates an integral whose upper limit is a variable. If $f$ is continuous on an interval containing $a$ and $g(x) = \int_a^x f(t)\,dt$, then: $$g'(x) = f(x)$$ NOTE (Which one is Part 1?): Textbooks disagree. Many (including Stewart) call the accumulation form Part 1 and the evaluation form Part 2; others swap the numbers or call one the Second Fundamental Theorem. The AP CED never numbers them: it files accumulation under Topic 6.4 and evaluation under Topic 6.7. Learn what each statement does, not the label. | Aspect | Evaluation form (Part 2) | Accumulation form (Part 1) | | --- | --- | --- | | CED topic | 6.7 (FUN-6.B.3) | 6.4 (FUN-6.B.2) | | What it does | Computes a definite integral | Differentiates an integral | | Limits | Both constant | A variable in a limit | | Output | A number | A function | | Core formula | $\int_a^b f(x)\,dx = F(b)-F(a)$ | $g'(x) = f(x)$ where $g(x)=\int_a^x f(t)\,dt$ | ### Part 2: turning an integral into a number Reach for the evaluation form whenever you need the value of a definite integral $\int_a^b f(x)\,dx$ and $f$ has an antiderivative you can find. This is the workhorse of Unit 6 and every integration application after it: area, average value, total change, volume. The method is two steps. First, find an antiderivative $F$ of the integrand (Topic 6.8 basic rules, or substitution in Topic 6.9). Second, compute $F(b) - F(a)$, usually written with a bracket: $$\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)$$ You never need the constant of integration here. If you used $F(x) + C$ instead of $F(x)$, the $+C$ appears in both $F(b)$ and $F(a)$ and cancels in the subtraction, so pick the simplest antiderivative. The theorem also requires $f$ continuous on the whole interval $[a,b]$: if the integrand blows up somewhere inside, this shortcut does not apply directly (that gap is what BC improper integrals in Topic 6.13 handle). The evaluation form is not obvious on its own; it works because of the accumulation form. Since $F(x) = \int_a^x f(t)\,dt$ is an antiderivative of $f$ (FUN-6.B.2), and any two antiderivatives of the same function differ only by a constant, evaluating any antiderivative at $b$ and $a$ and subtracting recovers the accumulated area from $a$ to $b$. That is the bridge between the two parts: Part 1 guarantees the antiderivative exists, Part 2 puts it to work. ### Part 1: the accumulation-function view The accumulation form is about a specific kind of function: one defined by an integral with a variable upper limit. $$g(x) = \int_a^x f(t)\,dt$$ Read $g(x)$ as the net area accumulated by $f$ from the fixed start $a$ out to $x$. The letter $t$ is just a placeholder (a dummy variable) for the integration; the real input is $x$, sitting in the upper limit. FUN-6.B.2 guarantees this $g$ is an antiderivative of $f$, and the accumulation form gives its derivative directly: $$g'(x) = f(x)$$ That is the whole point of Part 1: you can differentiate an integral without ever evaluating it. When the integrand has no elementary antiderivative, like $\sin(t^2)$ or $e^{-t^2}$, this is the only way to get the derivative at all. In an applied setting $g(x)$ reads as a running total. If $f(t)$ is a rate (liters per minute, meters per second), then $g(x)$ is the accumulated amount from time $a$ to time $x$, and $g'(x) = f(x)$ just says the instantaneous rate of that running total is the original rate. That interpretation is exactly Topic 6.5, where you analyze the behavior of $g$ from the graph of $f$. One extension is tested constantly: when the upper limit is itself a function $u(x)$ rather than plain $x$, the chain rule tacks on a factor of $u'(x)$. $$h(x) = \int_a^{u(x)} f(t)\,dt \Rightarrow h'(x) = f(u(x)) \cdot u'(x)$$ If the variable sits in the lower limit instead, swap the limits first and pick up a negative sign, because reversing the limits of a definite integral negates it (Topic 6.6). ### Deciding which part a problem wants The fastest tell is the limits of integration. Look there before you do anything else. Two constant limits, and the task is to produce a value (a number, an area, a total change, an average): that is the evaluation form. You will be finding an antiderivative and subtracting. A variable in a limit, usually paired with a definition like $g(x)$ built from an integral with $x$ in the upper limit, then a question about $g'$, about where $g$ is increasing or has a maximum, or about the concavity of $g$: that is the accumulation form. Any time a new function is built out of an integral and you are asked to differentiate or analyze it, Part 1 is the tool. - Numeric limits, asked for a value: evaluation form (Part 2). Antidifferentiate, then $F(b)-F(a)$. - Variable upper limit, asked for a derivative: accumulation form (Part 1). Replace $t$ with the limit, times its derivative. - Integrand you cannot antidifferentiate, but only a derivative is asked: accumulation form (Part 1). - Function defined as an integral, asked about increasing/decreasing or extrema: accumulation form (Part 1), then standard Unit 5 analysis on $g'$. ### Where students lose points - Dropping the chain rule factor. With a variable upper limit like $x^2$, the answer is $f(x^2)\cdot 2x$, not just $f(x^2)$. - Confusing the dummy variable. Inside $\int_a^x f(t)\,dt$ the integrand uses $t$; the $x$ appears only in the limit. Do not substitute $x$ for the integration variable. - Subtracting in the wrong order. It is always $F(\text{upper}) - F(\text{lower})$, that is $F(b) - F(a)$. Flipping the order flips the sign. - Ignoring continuity. The evaluation form needs $f$ continuous on $[a,b]$. A vertical asymptote inside the interval breaks it. - Forgetting the sign when the variable is the lower limit: $\dfrac{d}{dx}\int_x^b f(t)\,dt = -f(x)$. ### Worked examples #### Part 2: evaluate a definite integral Problem: Evaluate $\int_1^4 \left(2x + \dfrac{1}{\sqrt{x}}\right)\,dx$. 1. The limits are the constants $1$ and $4$ and the task is a value, so this is the evaluation form. First rewrite $\dfrac{1}{\sqrt{x}}$ as $x^{-1/2}$ to make antidifferentiation easy. 2. Antidifferentiate term by term. Since $\int 2x\,dx = x^2$ and $\int x^{-1/2}\,dx = \dfrac{x^{1/2}}{1/2} = 2\sqrt{x}$, an antiderivative is $F(x) = x^2 + 2\sqrt{x}$. 3. Evaluate at the upper limit: $F(4) = 4^2 + 2\sqrt{4} = 16 + 4 = 20$. 4. Evaluate at the lower limit: $F(1) = 1^2 + 2\sqrt{1} = 1 + 2 = 3$. 5. Subtract in order: $F(4) - F(1) = 20 - 3 = 17$. Answer: The integral equals 17. #### Part 1: differentiate an accumulation function Problem: Find $g'(x)$ for $g(x) = \int_2^x \sqrt{t^3+1}\,dt$, and $h'(x)$ for $h(x) = \int_0^{x^2} \sin(t^2)\,dt$. 1. Both define a function as an integral with a variable in the upper limit, so both call for the accumulation form. Notice neither integrand has an elementary antiderivative, so there is no way to evaluate first: Part 1 is the only route. 2. For $g$, the upper limit is plain $x$. Replace the integration variable $t$ with $x$ in the integrand: $g'(x) = \sqrt{x^3 + 1}$. 3. For $h$, the upper limit is $u(x) = x^2$, so the chain rule applies. First $u'(x) = 2x$. 4. Substitute $u(x) = x^2$ into the integrand $\sin(t^2)$, giving $\sin\left((x^2)^2\right) = \sin(x^4)$, then multiply by $u'(x)$: $h'(x) = \sin(x^4)\cdot 2x = 2x\sin(x^4)$. Answer: g'(x) = sqrt(x^3 + 1), and h'(x) = 2x sin(x^4). #### Both parts on one function Problem: Let $g(x) = \int_0^x (3t^2 - 4)\,dt$. (a) Use the accumulation form to find $g'(x)$. (b) Use the evaluation form to find a closed form for $g(x)$. (c) Confirm the two results agree. 1. (a) The upper limit is the plain variable $x$, so by the accumulation form replace $t$ with $x$ in the integrand: $g'(x) = 3x^2 - 4$. 2. (b) Here the integrand $3t^2 - 4$ does have an antiderivative, so the evaluation form gives a formula. An antiderivative is $t^3 - 4t$. 3. Apply $F(x) - F(0)$ with $F(t) = t^3 - 4t$: $g(x) = (x^3 - 4x) - (0^3 - 4\cdot 0) = x^3 - 4x$. 4. (c) Differentiate the closed form: the derivative of $x^3 - 4x$ is $3x^2 - 4$, which is exactly the $g'(x)$ from part (a). The two forms of the theorem agree, as they must. 5. This is why Part 1 is worth trusting: even when you could compute $g(x)$ in closed form, $g'(x) = 3x^2 - 4$ comes for free, telling you at once that $g$ is increasing wherever $3x^2 - 4 > 0$, that is for $|x| > \dfrac{2}{\sqrt{3}}$. Answer: g'(x) = 3x^2 - 4, g(x) = x^3 - 4x, and differentiating the closed form recovers 3x^2 - 4. ### FAQ Q: What is the difference between FTC Part 1 and Part 2? A: Part 1 (the accumulation form, CED Topic 6.4) differentiates an integral with a variable upper limit: the derivative of $\int_a^x f(t)\,dt$ is $f(x)$. Part 2 (the evaluation form, Topic 6.7) computes a definite integral as $F(b) - F(a)$ for an antiderivative $F$. One gives a function's derivative; the other gives a number. Q: Do I need the constant of integration for a definite integral? A: No. In $F(b) - F(a)$ any constant $C$ appears in both terms and cancels, so use the simplest antiderivative. The constant only matters for indefinite integrals (Topic 6.8). Q: When does the chain rule apply to the Fundamental Theorem? A: Whenever the variable limit is a function of $x$ rather than plain $x$. For $\int_a^{u(x)} f(t)\,dt$ the derivative is $f(u(x))\cdot u'(x)$. If it is just $\int_a^x$, then $u(x) = x$ and $u'(x) = 1$, so no extra factor appears. Q: Is the Fundamental Theorem of Calculus on both AB and BC? A: Yes. It sits in Unit 6, which carries 15 to 20 percent of both the AB and BC exams, and it underlies the integration applications in Units 7, 8, and 9 (BC). ## Riemann Sums: Left, Right, Midpoint, and Trapezoidal URL: https://www.calclearn.org/guides/riemann-sums Last updated: 2026-07-22 Answer first: A Riemann sum estimates a definite integral by adding rectangle or trapezoid areas; the left, right, midpoint, and trapezoidal rules differ only in where each strip takes its height. For an increasing function the left sum underestimates and the right overestimates; concavity decides midpoint and trapezoidal. ### When a Riemann sum is the right tool Reach for a Riemann sum whenever you need the area under a curve, or the total accumulated by a rate, but an exact antiderivative is out of reach: a table of measured data, a graph with no formula, or a function whose antiderivative you are not expected to find. This is Topics 6.2 and 6.3 in the CED, and it is where the definite integral is born. The plan never changes. Cut the interval $[a,b]$ into thin subintervals, build a rectangle or trapezoid over each one, and add the pieces. Narrower strips give a closer estimate. Every Riemann sum is a sum of products, and each product is one strip's height (a function value) times that strip's width. Once you see that structure you can build any of the four sums the exam asks for. The Riemann sum slider at /interactives/riemann-sum-slider lets you drag the number of subintervals up and watch the estimate tighten toward the true area, and the calculator at /calculators/riemann-sum-visualizer draws the strips for a function you type in, so you can check a hand computation against the picture. ### Four rules, one difference: where the height comes from Fix a partition of $[a,b]$ into $n$ subintervals with a uniform width $\Delta x = \frac{b-a}{n}$ and grid points $x_0, x_1, \dots, x_n$. All four methods multiply widths by heights and add. They disagree only about which height each strip uses. - Left sum: the height is the function value at the left edge of each strip. - Right sum: the height is the value at the right edge. - Midpoint sum: the height is the value at the middle of each strip. - Trapezoidal sum: no single height. Each strip is a trapezoid whose two parallel sides are the left and right function values, so its area uses their average. $$L_n = \Delta x\cdot\left[f(x_0) + f(x_1) + \cdots + f(x_{n-1})\right]$$ $$R_n = \Delta x\cdot\left[f(x_1) + f(x_2) + \cdots + f(x_n)\right]$$ $$M_n = \Delta x\cdot\left[f\!\left(\tfrac{x_0+x_1}{2}\right) + \cdots + f\!\left(\tfrac{x_{n-1}+x_n}{2}\right)\right]$$ $$T_n = \frac{\Delta x}{2}\cdot\left[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\right]$$ The trapezoidal formula is not a new idea; it is the average of the left and right sums, $T_n = \frac{L_n + R_n}{2}$. That is why every interior grid point carries a coefficient of 2: each one is the right edge of one strip and the left edge of the next, so it gets counted twice. When a table hands you unequal widths, drop the single $\Delta x$ and multiply each height, or each averaged pair, by its own strip width. ### Over or underestimate: the question the exam always asks Topic 6.2 leans hard on one skill: deciding, without computing the true value, whether an approximation is too big or too small. Two features of the function settle it, and each governs a different pair of methods. Monotonicity, whether the function is increasing or decreasing, controls the left and right sums. Concavity, whether the graph bends up or down, controls the midpoint and trapezoidal sums. For a rising function the left edge of every strip is its shortest side, so left rectangles sit under the curve and the left sum underestimates, while the right edges are tallest and the right sum overestimates. Flip both for a falling function. For concavity, picture the straight chord the trapezoid draws between two points on the curve: when the graph is concave up, that chord lies above the curve, so the trapezoidal sum overestimates. The midpoint rectangle behaves the opposite way: it has the same area as the trapezoid you would get from the tangent line at the midpoint, and for a concave-up curve that tangent line runs below the curve, so the midpoint sum underestimates. | Method | Underestimate when | Overestimate when | | --- | --- | --- | | Left sum | $f$ increasing | $f$ decreasing | | Right sum | $f$ decreasing | $f$ increasing | | Trapezoidal | $f$ concave down | $f$ concave up | | Midpoint | $f$ concave up | $f$ concave down | NOTE (Exam shortcut): For a function with a single concavity across the whole interval, the midpoint and trapezoidal errors point in opposite directions, so the exact integral is trapped between $M_n$ and $T_n$. That fact alone can answer a multiple-choice question without any arithmetic. ### Summation notation and the jump to the definite integral Topic 6.3 packs the whole construction into one symbol. A general Riemann sum is written as a sum of products over the subintervals of a partition. $$\sum_{i=1}^{n} f(x_i^{*})\cdot\Delta x_i$$ Here $\Delta x_i$ is the width of the $i$th subinterval and $x_i^{*}$ is any sample point inside it: its left end gives a left sum, its right end a right sum, its middle a midpoint sum. With a uniform width $\Delta x = \frac{b-a}{n}$, the definite integral is defined as the limit of these sums as the strips shrink to nothing. $$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^{*})\cdot\Delta x$$ For a continuous function the choice of sample point stops mattering in the limit, which is why left, right, and midpoint sums all converge to the same number (Topic 6.3). Worked Example 3 runs this limit by hand so you can watch the definite integral emerge from pure algebra, no antiderivative required. ### Mistakes that cost points - Miscounting grid points. $n$ subintervals need $n+1$ grid points. The left sum uses the first $n$ of them and drops the last; the right sum drops the first and keeps the last. - Forcing one width onto a table. Data with unequal spacing has no single $\Delta x$. Multiply each strip by its own width, or the arithmetic quietly smears. - Forgetting the 2's in the trapezoidal formula. Every interior grid point is shared by two trapezoids, so it appears twice; only the two outer endpoints carry a coefficient of 1. - Swapping the over/under rules. Increasing versus decreasing decides left and right sums. Concavity decides midpoint and trapezoidal. Judging a left sum by concavity is a classic trap. - Calling a fine partition exact. More strips shrink the error but never erase it. Only the limit as $n \to \infty$ gives the exact integral. ### Worked examples #### Left and right sums for a power function Problem: Estimate $\int_1^5 x^2\,dx$ using left and right Riemann sums with four subintervals of equal width, then decide whether each is an over- or underestimate. 1. The interval is $[1,5]$ with $n = 4$, so $\Delta x = \frac{5-1}{4} = 1$. The grid points are $x_0 = 1$, $x_1 = 2$, $x_2 = 3$, $x_3 = 4$, $x_4 = 5$, and $f(x) = x^2$. 2. Evaluate $f$ at every grid point: $f(1) = 1$, $f(2) = 4$, $f(3) = 9$, $f(4) = 16$, $f(5) = 25$. 3. The left sum uses the left endpoints $x_0$ through $x_3$: $L_4 = 1\cdot(1 + 4 + 9 + 16) = 30$. 4. The right sum uses the right endpoints $x_1$ through $x_4$: $R_4 = 1\cdot(4 + 9 + 16 + 25) = 54$. 5. Classify with monotonicity. Since $f'(x) = 2x > 0$ on $[1,5]$, the function is increasing, so the left sum underestimates and the right sum overestimates. As a check, the antiderivative gives $\int_1^5 x^2\,dx = \left[\frac{x^3}{3}\right]_1^5$. This evaluates to $\frac{125 - 1}{3} = \frac{124}{3}$, or about $41.33$, which sits between $30$ and $54$. Answer: $L_4 = 30$ (underestimate) and $R_4 = 54$ (overestimate); the exact value $\frac{124}{3} \approx 41.33$ lies between them. #### Trapezoidal sum from a table with unequal widths Problem: Water flows into a tank at a rate $R(t)$ liters per minute, measured at these times: $R(0) = 20$, $R(3) = 26$, $R(5) = 29$, $R(8) = 30$, and $R(12) = 35$. Use a trapezoidal sum with the four subintervals given by the table to approximate $\int_0^{12} R(t)\,dt$, and state what the result represents. 1. The partition is $0, 3, 5, 8, 12$, giving four subintervals of widths $3$, $2$, $3$, and $4$ minutes. Because the widths differ, handle each trapezoid separately instead of factoring out one $\Delta x$. 2. Each trapezoid area is the width times the average of its two heights: $\Delta t \cdot \frac{R(\text{left}) + R(\text{right})}{2}$. 3. Compute each piece. On $[0,3]$: $3 \cdot \frac{20 + 26}{2} = 69$. On $[3,5]$: $2 \cdot \frac{26 + 29}{2} = 55$. On $[5,8]$: $3 \cdot \frac{29 + 30}{2} = 88.5$. On $[8,12]$: $4 \cdot \frac{30 + 35}{2} = 130$. 4. Add the four trapezoids: $69 + 55 + 88.5 + 130 = 342.5$. Answer: About $342.5$ liters flow into the tank during the first $12$ minutes, since integrating a rate in liters per minute over minutes accumulates a total amount in liters. #### From a right Riemann sum to the definite integral Problem: Write the right Riemann sum for $\int_0^2 x^2\,dx$ with $n$ equal subintervals, then take the limit as $n \to \infty$ to evaluate the integral straight from the definition. 1. With $n$ equal strips on $[0,2]$, the width is $\Delta x = \frac{2}{n}$ and the right endpoints are $x_i = 0 + i\cdot\Delta x = \frac{2i}{n}$ for $i = 1, 2, \dots, n$. 2. Build the sum: $R_n = \sum_{i=1}^{n} f(x_i)\cdot\Delta x = \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^2 \cdot \frac{2}{n}$. 3. Simplify the general term: $\left(\frac{2i}{n}\right)^2 \cdot \frac{2}{n} = \frac{4i^2}{n^2}\cdot\frac{2}{n} = \frac{8i^2}{n^3}$, so $R_n = \frac{8}{n^3}\sum_{i=1}^{n} i^2$. 4. Apply the sum-of-squares formula $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$. Substituting gives $R_n = \frac{8}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}$, which simplifies to $\frac{4(n+1)(2n+1)}{3n^2}$. 5. Expand the numerator to get $R_n = \frac{4(2n^2 + 3n + 1)}{3n^2}$. Dividing each term by $n^2$ gives $\frac{4}{3}\left(2 + \frac{3}{n} + \frac{1}{n^2}\right)$. 6. Let $n \to \infty$. The terms $\frac{3}{n}$ and $\frac{1}{n^2}$ vanish, so $\lim_{n \to \infty} R_n = \frac{4}{3}\cdot 2$, which is $\frac{8}{3}$. Answer: $\int_0^2 x^2\,dx = \frac{8}{3}$, matching the antiderivative check $\left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}$. ### FAQ Q: Does the left or right Riemann sum overestimate? A: It depends on whether the function is increasing or decreasing, not on which sum you pick. For an increasing function the right sum overestimates and the left sum underestimates; for a decreasing function it is the reverse. If the function turns around inside the interval, split it at the turning point and judge each piece separately. Q: Is a midpoint Riemann sum an over- or underestimate? A: Concavity decides it. Where the graph is concave up, the midpoint sum underestimates; where it is concave down, the midpoint sum overestimates. This is the opposite of the trapezoidal rule, which overestimates when concave up and underestimates when concave down. Q: What is the difference between a Riemann sum and a definite integral? A: A Riemann sum is a finite estimate: a fixed number of strips whose areas you add. The definite integral $\int_a^b f(x)\,dx$ is the exact value those sums approach as the strips get infinitely thin, $\lim_{n \to \infty} \sum f(x_i^{*})\cdot\Delta x$. Every Riemann sum stays an approximation until you take the limit. Q: Why is the trapezoidal rule usually more accurate than left or right sums? A: A trapezoid follows the straight line between two points on the curve, while a rectangle uses only one endpoint's height, so the trapezoid tracks a sloped function much more closely. For the same number of subintervals it is the average of the left and right sums, $T_n = \frac{L_n + R_n}{2}$, which cancels much of the error the one-sided sums carry in opposite directions. ## Disk vs Washer vs Shell: Choosing the Volume Method URL: https://www.calclearn.org/guides/disk-washer-shell Last updated: 2026-07-22 Answer first: Slice the region perpendicular to the axis of revolution: no gap between the slice and the axis means the disk method; a gap means the washer method. On the AP exam only disk and washer are tested (CED 8.9-8.12); the shell method is a college topic, not on the exam. ### Draw the axis, then draw one slice The fastest way to choose a volume method is to draw two things and read the picture. First draw the axis of revolution. Then draw one representative slice cut perpendicular to that axis. When you spin that perpendicular slice around the axis it sweeps out a flat circular piece, and whether that piece is a solid disk or a ring with a hole punched out is the whole decision. This is what Topics 8.9 through 8.12 are testing. If the slice runs from the axis straight out to the boundary with nothing in between, revolving it produces a solid disk. If there is a gap between the axis and the region, so the slice starts out away from the axis, revolving it produces a washer: a disk with a smaller disk removed from the center. Disk or washer comes down to one question: is there a gap between the region and the axis? NOTE (Decision rule): Slice perpendicular to the axis of revolution. No gap between the region and the axis means the disk method. A gap means the washer method. That single test decides every AP volume-of-revolution problem. ### Disk or washer: run the gap test Three situations cover almost every exam problem, and the gap test sorts them instantly. When one boundary of the region is the axis itself, say the region sits between a curve and the $x$-axis and you revolve around the $x$-axis, the slice starts right at the axis. No gap, so it is a disk. When the region is trapped between two curves that both stay off the axis and you revolve around that axis, every slice starts away from the axis at the nearer curve and ends at the farther curve. That is a gap, so it is a washer, and the two curves supply the inner and outer radii. The third case is the one students miss: the region touches the axis in its original position, but the problem revolves it around a different line. Moving the axis can open a gap where there was none. The same region that gives a disk around the $x$-axis gives a washer around the line $y = 2$, because the slice no longer starts at the axis. Always run the gap test against the axis the problem actually names, which is the whole point of Topics 8.10 and 8.12. A quick check before you commit: label the nearer and farther boundaries as seen from the axis. If the nearer boundary is the axis line itself, the inner radius is zero and the washer collapses to a disk, which confirms the no-gap case. If the nearer boundary is a curve, that curve supplies the inner radius (measure the distance from the axis to it) and you are in washer territory. ### The formulas, and why the radius is always a distance Both methods integrate the area of the circular slice. A disk of radius $R$ has area $\pi R^2$. A washer with outer radius $R_{\text{out}}$ and inner radius $R_{\text{in}}$ has area $\pi\left(R_{\text{out}}^2 - R_{\text{in}}^2\right)$. Multiply by the slice thickness and add up with an integral. $$V = \pi \int_a^b \left[R(x)\right]^2 \,dx$$ $$V = \pi \int_a^b \left(\left[R_{\text{out}}(x)\right]^2 - \left[R_{\text{in}}(x)\right]^2\right) \,dx$$ Here is the idea that prevents most setup errors: the radius is a distance from the axis to a boundary, not simply a function value. When you revolve around the $x$-axis the radius equals the function value, which hides the distinction. The instant the axis moves to a line such as $y = 2$ or $x = 3$, you must write the radius as the axis position minus the curve, or the curve minus the axis position, whichever comes out positive. For a washer the outer radius reaches the boundary farther from the axis and the inner radius reaches the nearer one. The slice thickness tells you the variable of integration. A horizontal axis, the $x$-axis or any line $y = k$, produces vertical slices of thickness $dx$, so you integrate in $x$. A vertical axis, the $y$-axis or any line $x = k$, produces horizontal slices of thickness $dy$, so you integrate in $y$ and every boundary must be rewritten as a function of $y$. Match the variable to the direction of the axis and the rest of the setup follows. Finding the limits is the same intersection work you did for area between curves in Topics 8.4 through 8.6. | Axis of revolution | Region touches the axis? | Method | Integrate in | | --- | --- | --- | --- | | Horizontal: $x$-axis or $y = k$ | Yes (no gap) | Disk | $x$, thickness $dx$ | | Horizontal: $x$-axis or $y = k$ | No (a gap) | Washer | $x$, thickness $dx$ | | Vertical: $y$-axis or $x = k$ | Yes (no gap) | Disk | $y$, thickness $dy$ | | Vertical: $y$-axis or $x = k$ | No (a gap) | Washer | $y$, thickness $dy$ | ### Where the shell method fits (and why the AP exam skips it) The shell method slices parallel to the axis instead of perpendicular to it. Revolving a parallel slice sweeps out a thin cylindrical shell, like the paper label on a soup can, with volume $2\pi$ times radius times height times thickness. It is a genuine and useful technique, and you will most likely meet it in a college calculus course. $$V = 2\pi \int_a^b (\text{radius})(\text{height}) \,dx$$ NOTE (Not on the AP exam): The shell method is not part of AP Calculus AB or BC. The CED covers only the disk method (Topics 8.9 and 8.10) and the washer method (Topics 8.11 and 8.12). Every exam volume problem is solvable with disk or washer, so you never need shells on a free-response question. Learn it if a college course or your teacher asks for it, but do not reach for it on the AP exam. Shells exist because a perpendicular slice sometimes forces you to solve a curve for the awkward variable or to split the region into two separate integrals, while a parallel slice keeps everything in a single clean integral. AP problems are written so that disk or washer always stays workable, which is exactly why the exam can leave shells out without limiting what it can ask. ### Mistakes that cost points - Forgetting to square before integrating. The integrand is the radius squared, not the radius. Square first, then antidifferentiate. - Subtracting the radii instead of their squares. A washer is $\pi\left(R_{\text{out}}^2 - R_{\text{in}}^2\right)$, never $\pi\left(R_{\text{out}} - R_{\text{in}}\right)^2$. Those two expressions are not equal. - Using the bare function value as the radius after the axis has moved. Around $y = 2$ the radius is $2 - (\text{curve})$ or $(\text{curve}) - 2$, not the curve by itself. - Mismatching the variable and the axis. Around a vertical line you must integrate in $y$, with every boundary rewritten as a function of $y$. - Skipping the intersection step. Find where the boundaries meet first; those points are your limits of integration. ### Worked examples #### Disk method around the x-axis Problem: The region bounded by $y = \sqrt{x}$, the $x$-axis, and the line $x = 4$ is revolved around the $x$-axis. Find the volume. 1. Draw the axis of revolution (the $x$-axis) and a representative vertical slice, which is perpendicular to that axis. The slice runs from the $x$-axis up to the curve $y = \sqrt{x}$, so it touches the axis with no gap. A perpendicular slice that touches the axis sweeps out a solid disk, so this is the disk method. 2. The radius of each disk is the distance from the axis to the curve: $R(x) = \sqrt{x}$. Because the axis is horizontal, the slices have thickness $dx$ and you integrate in $x$, from $x = 0$ to $x = 4$. 3. Set up the disk integral: $V = \pi \int_0^4 \left(\sqrt{x}\right)^2 \,dx$. 4. Simplify the integrand: $\left(\sqrt{x}\right)^2 = x$, so $V = \pi \int_0^4 x \,dx$. 5. Antidifferentiate: $\int x \,dx = \frac{x^2}{2}$, giving $V = \pi \left[\frac{x^2}{2}\right]_0^4$. 6. Evaluate: $V = \pi \left(\frac{16}{2} - 0\right) = 8\pi$. Answer: $V = 8\pi$ cubic units. #### Washer method around the x-axis Problem: The region enclosed by $y = 2x$ and $y = x^2$ is revolved around the $x$-axis. Find the volume. 1. Find where the curves meet: set $2x = x^2$, so $x^2 - 2x = 0$ and $x(x - 2) = 0$, giving $x = 0$ and $x = 2$. The region lives on $[0, 2]$. 2. Decide which curve is on top. At $x = 1$, $2x = 2$ and $x^2 = 1$, so $y = 2x$ is the upper boundary and $y = x^2$ is the lower boundary. 3. Draw the $x$-axis and a vertical slice. The lower boundary $y = x^2$ sits above the axis (except at the endpoint $x = 0$), so there is a gap between the region and the axis. A perpendicular slice with a gap sweeps out a washer, so this is the washer method. 4. Identify the radii as distances from the $x$-axis. The outer radius reaches the farther curve: $R_{\text{out}} = 2x$. The inner radius reaches the nearer curve: $R_{\text{in}} = x^2$. 5. Set up the washer integral: $V = \pi \int_0^2 \left[(2x)^2 - (x^2)^2\right] \,dx$. 6. Expand each square: $(2x)^2 = 4x^2$ and $(x^2)^2 = x^4$, so $V = \pi \int_0^2 \left(4x^2 - x^4\right) \,dx$. 7. Antidifferentiate: $V = \pi \left[\frac{4x^3}{3} - \frac{x^5}{5}\right]_0^2$. 8. Evaluate at $x = 2$: $\frac{4(8)}{3} - \frac{32}{5} = \frac{32}{3} - \frac{32}{5}$. 9. Use a common denominator of $15$: $\frac{32}{3} = \frac{160}{15}$ and $\frac{32}{5} = \frac{96}{15}$, so the difference is $\frac{160 - 96}{15} = \frac{64}{15}$. Answer: $V = \frac{64\pi}{15}$ cubic units. #### Same region, new axis: a disk becomes a washer Problem: Take the same region as Example 1 (bounded by $y = \sqrt{x}$, the $x$-axis, and $x = 4$), but revolve it around the horizontal line $y = 2$. Find the volume. 1. Nothing about the region changed, only the axis moved. Draw the line $y = 2$ and a vertical slice. Since $\sqrt{x} \le 2$ on $[0, 4]$, the whole region sits below the axis, and the slice no longer touches $y = 2$ except at the single point $x = 4$. The gap means this is now a washer, not a disk. 2. Measure each radius as a distance from the axis $y = 2$. The boundary farther from the axis is the $x$-axis, where $y = 0$: $R_{\text{out}} = 2 - 0 = 2$. The nearer boundary is the curve $y = \sqrt{x}$: $R_{\text{in}} = 2 - \sqrt{x}$. 3. Set up the washer integral (horizontal axis, so thickness $dx$): $V = \pi \int_0^4 \left[2^2 - \left(2 - \sqrt{x}\right)^2\right] \,dx$. 4. Expand the inner square: $\left(2 - \sqrt{x}\right)^2 = 4 - 4\sqrt{x} + x$. 5. Subtract inside the brackets: $4 - \left(4 - 4\sqrt{x} + x\right) = 4\sqrt{x} - x$, so $V = \pi \int_0^4 \left(4\sqrt{x} - x\right) \,dx$. 6. Antidifferentiate. Write $\sqrt{x} = x^{1/2}$, so $\int 4x^{1/2} \,dx = 4 \cdot \frac{x^{3/2}}{3/2} = \frac{8}{3}x^{3/2}$ and $\int x \,dx = \frac{x^2}{2}$. Then $V = \pi \left[\frac{8}{3}x^{3/2} - \frac{x^2}{2}\right]_0^4$. 7. Evaluate at $x = 4$: $x^{3/2} = 4^{3/2} = 8$, so $\frac{8}{3}(8) = \frac{64}{3}$, and $\frac{x^2}{2} = \frac{16}{2} = 8 = \frac{24}{3}$. 8. Combine: $\frac{64}{3} - \frac{24}{3} = \frac{40}{3}$. Answer: $V = \frac{40\pi}{3}$ cubic units. ### FAQ Q: Is the shell method on the AP Calculus exam? A: No. AP Calculus AB and BC test only the disk method (Topics 8.9 and 8.10) and the washer method (Topics 8.11 and 8.12). The shell method is common in college calculus but is never required on the AP exam, and every exam volume-of-revolution problem can be done with disk or washer. Q: How do I know whether to integrate with respect to x or y? A: Look at the axis of revolution. A horizontal axis (the $x$-axis or a line $y = k$) gives vertical slices of thickness $dx$, so you integrate in $x$. A vertical axis (the $y$-axis or a line $x = k$) gives horizontal slices of thickness $dy$, so you integrate in $y$ and rewrite every boundary as a function of $y$. Q: What is the difference between the disk and washer method? A: Both revolve a perpendicular slice around the axis. The disk method applies when the slice touches the axis with no gap, giving a solid disk of area $\pi R^2$. The washer method applies when there is a gap between the region and the axis, giving a ring of area $\pi\left(R_{\text{out}}^2 - R_{\text{in}}^2\right)$. The gap is the only thing that decides between them. ## Area Between Two Curves: Setup, dx vs dy, Examples URL: https://www.calclearn.org/guides/area-between-curves Last updated: 2026-07-22 Answer first: To find the area between two curves, integrate top minus bottom with respect to x (or right minus left with respect to y) between the intersection points. If the curves cross inside the interval, split the integral where they swap and add the pieces so each difference stays positive. ### The setup: upper boundary minus lower The area of a region enclosed by two curves is the integral of the upper boundary minus the lower boundary. When both curves are written as functions of $x$, that is top minus bottom, integrated across the interval where the region lives: $$A = \int_a^b \bigl[\, f(x) - g(x) \,\bigr]\,dx, \qquad f(x) \ge g(x) \text{ on } [a,b]$$ Slice the region into thin vertical strips. Each strip has height $f(x) - g(x)$, the gap between the curves, and width $dx$; multiplying gives the area of one strip, and the integral adds up every strip from $x = a$ to $x = b$. That is the same Riemann-sum idea from Unit 6, now measuring the space between two graphs instead of the space under one. This is Topic 8.4. One rule keeps you out of trouble: subtract in whatever order makes the height positive. Area is never negative, so the curve that is physically higher on the interval goes first. If you are unsure which is higher, plug a single test value from between the bounds into both functions and compare the outputs. ### dx or dy: which way do the strips run? The choice between integrating in $x$ or $y$ is a choice about strip direction, and it is the first real decision this topic asks you to make. Vertical strips give top minus bottom and integrate with respect to $x$. Horizontal strips give right minus left and integrate with respect to $y$ (Topic 8.5): $$A = \int_c^d \bigl[\, p(y) - q(y) \,\bigr]\,dy, \qquad p(y) \ge q(y) \text{ on } [c,d]$$ Two signals tell you to switch to $dy$. The first is the form of the curves: if a boundary is given as $x = f(y)$, such as $x = y^2$, horizontal strips read the picture directly. The second is what a vertical cut hits: if a single vertical strip would meet one curve on part of the region and a different curve elsewhere, integrating in $x$ forces you to break the region into several integrals, while a horizontal strip might sweep the whole region in one. Pick the direction that keeps a single pair of boundaries. | Feature | Vertical strips (dx) | Horizontal strips (dy) | | --- | --- | --- | | Strip runs | up and down | left and right | | You subtract | top minus bottom | right minus left | | Bounds are | x-values a to b | y-values c to d | | Best when | curves are y = f(x) | curves are x = f(y) | ### Finding the limits: where the curves meet The bounds of the integral are the points where the two curves intersect. Set the two expressions equal and solve. If you are integrating in $x$, set $f(x) = g(x)$ and solve for $x$; those $x$-values are $a$ and $b$. If you are integrating in $y$, set the two $x$-expressions equal and solve for $y$. On the AP exam this splits by section. On the no-calculator parts the intersections are built to factor cleanly, so a quadratic like $x^2 - x - 2 = 0$ factors to $(x - 2)(x + 1)$ and hands you the bounds. On the calculator-active parts you are expected to use the intersect feature, carry several decimal places, and integrate numerically. Either way, never assume the region starts at the $y$-axis or at $x = 0$; find the actual crossing points first. ### When the curves cross (Topic 8.6) If the two curves intersect at more than two points, the upper and lower boundaries trade places somewhere inside the region. A single subtraction no longer works for the whole interval, because on one piece $f$ is on top and on another piece $g$ is. Topic 8.6 handles exactly this case: split the interval at every interior crossing, subtract in the correct order on each piece, and add the results. $$A = \int_a^b \left| f(x) - g(x) \right|\,dx$$ The absolute value is the compact way to say the same thing: it automatically takes the positive height on every piece. To actually evaluate it you still split at the crossings and drop the bars with the right sign on each subinterval, so the work is identical. Finding those crossings is the whole game, so solve $f(x) = g(x)$ across the full domain, not just near the ends: a factor like $x(x - 1)(x + 1)$ reveals all three roots rather than two. $$\int_{-1}^{1} (x - x^3)\,dx = 0$$ NOTE (Do not integrate straight through a crossing): Without the absolute value or a split, the signed areas partly cancel and your answer comes out too small, sometimes zero. For $y = x$ and $y = x^3$ on $[-1, 1]$ the honest area is $\frac{1}{2}$, but integrating straight through gives $0$ (shown above) because the two halves cancel exactly. ### Pitfalls that cost points - Subtracting in the wrong order. Bottom minus top gives a negative number; if your area comes out negative, you flipped the difference. - Skipping a crossing. If the curves meet three or more times, integrating straight through makes the signed areas cancel. Split at every interior intersection. - Forcing $dx$ on a sideways region. When the boundaries are $x = f(y)$, horizontal strips and a $dy$ integral are usually one clean integral instead of two. - Using the wrong bounds. The limits are the intersection points of the two curves, not the $x$-intercepts of either curve alone. - Rushing the calculator setup on Section II. On calculator-active free response, write the exact integral first, then evaluate numerically; you earn the setup point even if the arithmetic slips. ### Worked examples #### Area with respect to x (Topic 8.4) Problem: Find the area of the region enclosed by $y = x + 2$ and $y = x^2$. 1. Find where the curves meet. Set $x + 2 = x^2$, which gives $x^2 - x - 2 = 0$, so $(x - 2)(x + 1) = 0$ and $x = -1$ or $x = 2$. These are the limits of integration. 2. Decide which curve is on top. Test $x = 0$: the line gives $0 + 2 = 2$ and the parabola gives $0$, so the line $y = x + 2$ is the upper boundary. The strips are vertical, so integrate with respect to $x$. 3. Set up top minus bottom: $A = \int_{-1}^{2} \bigl[(x + 2) - x^2\bigr]\,dx$. 4. Antidifferentiate term by term: $\int (x + 2 - x^2)\,dx = \frac{x^2}{2} + 2x - \frac{x^3}{3}$. 5. Evaluate at the upper limit $x = 2$: $\frac{4}{2} + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}$. 6. Evaluate at the lower limit $x = -1$: $\frac{1}{2} - 2 + \frac{1}{3} = -\frac{7}{6}$. 7. Subtract the two values: $\frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$. Answer: $A = \frac{9}{2}$ square units. #### Area with respect to y (Topic 8.5) Problem: Find the area of the region enclosed by $x = y^2$ and $x = 2y + 3$. 1. Find the intersections. Set $y^2 = 2y + 3$, giving $y^2 - 2y - 3 = 0$, so $(y - 3)(y + 1) = 0$ and $y = -1$ or $y = 3$. 2. Look at the region. The parabola $x = y^2$ opens to the right and the line $x = 2y + 3$ sits to its right; test $y = 0$, where the parabola gives $x = 0$ and the line gives $x = 3$. Horizontal strips run from the parabola on the left to the line on the right. 3. Choose $dy$. Solving $x = y^2$ for $y$ splits into $y = \pm\sqrt{x}$, so vertical strips would need two integrals, while horizontal strips need only one. Integrate right minus left with respect to $y$. 4. Set up the integral: $A = \int_{-1}^{3} \bigl[(2y + 3) - y^2\bigr]\,dy$. 5. Antidifferentiate: $\int (2y + 3 - y^2)\,dy = y^2 + 3y - \frac{y^3}{3}$. 6. Evaluate at $y = 3$: $9 + 9 - 9 = 9$. 7. Evaluate at $y = -1$: $1 - 3 + \frac{1}{3} = -\frac{5}{3}$. 8. Subtract: $9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{32}{3}$. Answer: $A = \frac{32}{3}$ square units. #### Curves that cross (Topic 8.6) Problem: Find the total area of the region enclosed between $y = x^3$ and $y = x$. 1. Find every intersection. Set $x^3 = x$, so $x^3 - x = 0$ and $x(x - 1)(x + 1) = 0$, giving $x = -1$, $x = 0$, and $x = 1$. Three intersection points mean the curves cross, so this is a Topic 8.6 region. 2. Determine the order on each piece. On $(-1, 0)$ test $x = -\frac{1}{2}$: $x^3 = -\frac{1}{8}$ sits above $x = -\frac{1}{2}$, so $y = x^3$ is on top. On $(0, 1)$ test $x = \frac{1}{2}$: $x = \frac{1}{2}$ sits above $x^3 = \frac{1}{8}$, so $y = x$ is on top. 3. Split the area at $x = 0$ and subtract in the correct order on each piece: $A = \int_{-1}^{0} (x^3 - x)\,dx + \int_{0}^{1} (x - x^3)\,dx$. 4. First integral: $\int (x^3 - x)\,dx = \frac{x^4}{4} - \frac{x^2}{2}$. At $x = 0$ this is $0$; at $x = -1$ it is $\frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$. The piece equals $0 - \left(-\frac{1}{4}\right) = \frac{1}{4}$. 5. Second integral: $\int (x - x^3)\,dx = \frac{x^2}{2} - \frac{x^4}{4}$. At $x = 1$ this is $\frac{1}{2} - \frac{1}{4} = \frac{1}{4}$; at $x = 0$ it is $0$. The piece equals $\frac{1}{4}$. 6. Add the pieces: $A = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$. Answer: $A = \frac{1}{2}$ square units. ### FAQ Q: Do you always subtract top minus bottom? A: When you integrate with respect to $x$, yes: the upper curve minus the lower curve. When you integrate with respect to $y$, the parallel rule is right minus left. In both cases you subtract so the height is positive, because area cannot be negative. Q: When do you integrate with respect to y instead of x? A: Switch to $dy$ when the boundaries are given as $x = f(y)$, or when a single vertical strip would hit different curves in different parts of the region. Horizontal strips often turn a two-integral problem into one. Q: What do you do when the curves cross in the middle? A: Split the interval at every interior intersection. On each piece, subtract the curve that is on top there, then add the pieces. Equivalently, integrate the absolute value of the difference. This is Topic 8.6. Q: Is area between curves on the calculator or non-calculator part? A: Both. Non-calculator problems use curves whose intersections factor cleanly. Calculator-active free response expects you to find intersections with the intersect feature and evaluate the integral numerically. ## Product Rule vs Quotient Rule: How and When to Use Each URL: https://www.calclearn.org/guides/product-quotient-rule Last updated: 2026-07-22 Answer first: Use the product rule when two functions are multiplied and the quotient rule when one is divided by another. The product rule gives f'g + fg'; the quotient rule gives (gf' - fg')/g^2. Before using the quotient rule, check whether rewriting the fraction as a product or a sum of powers is faster. ### Which rule does the function need? Before differentiating anything, look at the outermost operation in the function, the last step you would perform if you plugged in a number. If two nonconstant functions are multiplied, use the product rule (Topic 2.8). If one function is divided by another, use the quotient rule (Topic 2.9). That one question settles the choice, and it matters because the two rules are built differently and are not interchangeable at a glance. Two checks come first. Is either piece just a constant? A constant times a function is the constant multiple rule from Topic 2.6, so the derivative of $7\cos x$ is $-7\sin x$ with no product rule involved. Can the expression be rewritten so the multiplication or division disappears? A fraction whose denominator is a single power can be split into separate terms and finished with the power rule alone. Running these checks first is the method-selection habit that keeps you off the longest, most error-prone path. NOTE (Decision rule): Product of two functions goes to the product rule. One function over another goes to the quotient rule. But first ask whether a constant multiple or a quick rewrite makes both rules unnecessary. | | Product rule | Quotient rule | | --- | --- | --- | | Trigger | two functions multiplied | one function divided by another | | Formula | $f'g + fg'$ | $\dfrac{gf' - fg'}{g^2}$ | | Term order | addition, order is free | subtraction, order is fixed | | Common slip | writing $f'g'$ | flipping the numerator terms | ### The product rule (Topic 2.8) The product rule says the derivative of a product is the derivative of the first factor times the second, plus the first factor times the derivative of the second. $$\frac{d}{dx}\left[f(x)g(x)\right] = f'(x)g(x) + f(x)g'(x)$$ Because the two terms are added, their order does not matter: you can write $f(x)g'(x) + f'(x)g(x)$ and the answer is identical. The mistake to avoid is assuming the derivative of a product is the product of the derivatives. The expression $f'(x)g'(x)$ is not the derivative of $f(x)g(x)$; each function has to be paired with the other's derivative. Worked example 1 applies the rule to $x^2\sin x$. In practice you often combine the product rule with other rules from Unit 2. Each factor is differentiated with whatever rule fits it, a power rule here, a trig or exponential rule there, and those individual derivatives drop into the product-rule template. The structure of the rule does not change; only the pieces you feed it do. ### The quotient rule and low-d-high (Topic 2.9) The derivative of a quotient is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the denominator squared. $$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{\left[g(x)\right]^{2}}$$ A popular memory aid is low-d-high: low d-high minus high d-low, over the square of what is below. Read high as the numerator, low as the denominator, and d as take the derivative of. Used honestly, the mnemonic has one catch worth stating plainly. The subtraction is not symmetric. With the product rule you add two terms and the order is free; here the term holding the derivative of the numerator must come first, and swapping the two terms flips the sign of the entire derivative. The denominator squared applies to the bottom function only, not to the whole fraction. It helps to see where the mnemonic comes from. Writing $\dfrac{f}{g}$ as $f \cdot g^{-1}$ and differentiating with the product rule and chain rule produces exactly the quotient rule after you combine terms over a common denominator. That is why, whenever the order slips your mind, the product-rule form is a reliable fallback. NOTE (Order matters): Low-d-high comes before high-d-low. If you cannot remember which term leads, rewrite the quotient as a product with a negative exponent and use the product rule instead, which never has an order trap. ### When rewriting skips the quotient rule The quotient rule is never wrong, but it is often the slow way. Before you apply it, check for three situations where a rewrite is cleaner and less error-prone. - The denominator is a single term. Split the fraction into separate fractions and reduce each with the power rule, as in worked example 3. A denominator like $x^2$ or $3x^5$ almost always signals this shortcut. - The denominator is a constant. Then the whole function is a constant multiple: the derivative of $\dfrac{x^3 - x}{4}$ is $\dfrac{3x^2 - 1}{4}$, using Topic 2.6. - The numerator is a constant. Write the function as that constant times the denominator to the power $-1$ and differentiate with the chain rule, which is usually shorter than the quotient rule. None of these needs the quotient rule at all. Recognizing them is the whole point: the exam rewards choosing the shortest correct path, not grinding every fraction through the same formula. ### Common mistakes to avoid Most lost points on these topics come from a small set of predictable slips rather than from not knowing the rules. On multiple-choice questions the wrong answers are usually built from exactly these errors, so a graded response that matches one of the traps below is a signal to recheck rather than move on. Watch for each of the following. - Writing $f'(x)g'(x)$ for the derivative of a product. The product rule mixes each function with the other function's derivative; it never multiplies the two derivatives. - Flipping the subtraction order in the quotient rule. Low-d-high comes first; reversing the terms negates the whole answer. - Forgetting to square the denominator, or squaring the entire fraction instead of just the bottom function. - Reaching for the quotient rule when the denominator is a single power or a constant, where splitting or a constant multiple is faster. - Missing a hidden chain rule. If the numerator or denominator is itself a composition, such as $\sin(3x)$, its derivative needs the chain rule before it goes into the product or quotient rule. ### Worked examples #### Product rule with a trig factor Problem: Differentiate $f(x) = x^2 \sin x$. 1. The outer operation is multiplication of $x^2$ and $\sin x$, so this is a product rule problem. Let the first factor be $x^2$ and the second be $\sin x$. 2. Differentiate each factor separately: the derivative of $x^2$ is $2x$, and the derivative of $\sin x$ is $\cos x$ (Topic 2.7). 3. Apply the product rule, derivative of the first times the second plus the first times the derivative of the second: $f'(x) = (2x)(\sin x) + (x^2)(\cos x)$. 4. Write the terms in a clean order: $f'(x) = 2x\sin x + x^2\cos x$. Nothing cancels, so this is fully simplified. Answer: $f'(x) = 2x\sin x + x^2\cos x$ #### Quotient rule step by step Problem: Differentiate $f(x) = \dfrac{2x+1}{x^2+3}$. 1. The outer operation is division, and the denominator $x^2+3$ is not a single power, so the quotient rule applies. Set high to $2x+1$ and low to $x^2+3$. 2. Differentiate each piece: the derivative of the high is $2$, and the derivative of the low is $2x$. 3. Apply low-d-high minus high-d-low over low squared: $f'(x) = \dfrac{(x^2+3)(2) - (2x+1)(2x)}{(x^2+3)^2}$. 4. Expand the numerator: $(x^2+3)(2) = 2x^2 + 6$ and $(2x+1)(2x) = 4x^2 + 2x$. 5. Subtract, keeping the order fixed: $2x^2 + 6 - (4x^2 + 2x) = 2x^2 + 6 - 4x^2 - 2x = -2x^2 - 2x + 6$. 6. So $f'(x) = \dfrac{-2x^2 - 2x + 6}{(x^2+3)^2}$. Factoring $-2$ from the numerator gives the equivalent form $\dfrac{-2(x^2 + x - 3)}{(x^2+3)^2}$. Answer: $f'(x) = \dfrac{-2x^2 - 2x + 6}{(x^2+3)^2}$ #### Rewriting to skip the quotient rule Problem: Differentiate $f(x) = \dfrac{x^3 + 4x}{x^2}$. 1. The denominator is a single power, $x^2$, so before reaching for the quotient rule, split the fraction into separate terms: $f(x) = \dfrac{x^3}{x^2} + \dfrac{4x}{x^2}$. 2. Reduce each term using exponent rules: $\dfrac{x^3}{x^2} = x$ and $\dfrac{4x}{x^2} = 4x^{-1}$. So $f(x) = x + 4x^{-1}$. 3. Differentiate term by term with the power rule (Topic 2.5): the derivative of $x$ is $1$, and the derivative of $4x^{-1}$ is $-4x^{-2}$. 4. Combine: $f'(x) = 1 - 4x^{-2} = 1 - \dfrac{4}{x^2}$. 5. As a check, the quotient rule gives $\dfrac{x^2(3x^2+4) - (x^3+4x)(2x)}{x^4} = \dfrac{x^4 - 4x^2}{x^4} = 1 - \dfrac{4}{x^2}$, the same answer with more algebra. Answer: $f'(x) = 1 - \dfrac{4}{x^2}$ ### FAQ Q: Is there a product rule for three functions? A: Yes. For a product of three functions the derivative is $f'gh + fg'h + fgh'$: differentiate one factor at a time and keep the other two unchanged. The same pattern extends to any number of factors. Q: Can I use the product rule instead of the quotient rule? A: Always. Rewrite $\dfrac{f}{g}$ as $f \cdot g^{-1}$ and use the product rule together with the chain rule. Combining the result over a common denominator gives exactly the quotient rule, so it is a valid fallback if you forget the quotient formula. Q: What is the difference between the product rule and the chain rule? A: The product rule is for two functions multiplied together, such as $x^2\sin x$. The chain rule is for one function composed inside another, such as $\sin(x^2)$. Check whether the pieces are multiplied or nested before choosing. Q: Do I need to memorize the quotient rule for the AP exam? A: Yes. The AP Calculus AB and BC exams do not provide a formula sheet, so you need the quotient rule memorized, or be ready to rebuild it from the product rule with a negative exponent. ## Derivatives of Trig Functions: Table and Memory Tricks URL: https://www.calclearn.org/guides/trig-derivatives Last updated: 2026-07-22 Answer first: The six trig derivatives are: sin x gives cos x, cos x gives -sin x, tan x gives sec^2 x, cot x gives -csc^2 x, sec x gives sec x tan x, and csc x gives -csc x cot x. The memory trick: every co-function (cosine, cotangent, cosecant) picks up a negative sign in its derivative. ### The six trig derivatives at a glance Every trig derivative you need in AP Calculus comes from this table. Sine and cosine (Topic 2.7) are the two you truly memorize; tangent, cotangent, secant, and cosecant (Topic 2.10) all follow from those two through the quotient rule, so you can rebuild them in a few seconds if a formula slips. Recognition here is simple: if you see one of the six named trig functions with nothing composed inside it, apply the matching row directly. | Function $f(x)$ | Derivative $f'(x)$ | | --- | --- | | $\sin x$ | $\cos x$ | | $\cos x$ | $-\sin x$ | | $\tan x$ | $\sec^2 x$ | | $\cot x$ | $-\csc^2 x$ | | $\sec x$ | $\sec x \tan x$ | | $\csc x$ | $-\csc x \cot x$ | The recognition question is not which technique, it is which row, and then whether anything else is happening around the trig function. A bare $\sin x$ or $\sec x$ takes the table entry as is. The moment an inner function appears, such as $\sin(3x)$ or $\sec(x^2)$, you still start from the table but multiply by the inner derivative through the chain rule (Topic 3.1). Keep those two steps separate in your head. ### The co-function negative pattern The fastest way to hold all six formulas in memory is one sign rule: every co-function differentiates to a negative. The co-functions are the three whose names start with co, meaning cosine, cotangent, and cosecant, and their derivatives all carry a minus sign. The other three, sine, tangent, and secant, stay positive. That single observation cuts what you have to remember roughly in half, because the six formulas pair up. - $\sin x$ and $\cos x$ are a pair: the derivative of $\sin x$ is $+\cos x$, and the derivative of $\cos x$ is $-\sin x$. - $\tan x$ and $\cot x$ are a pair: $\tan x$ gives $+\sec^2 x$, and $\cot x$ gives $-\csc^2 x$. - $\sec x$ and $\csc x$ are a pair: $\sec x$ gives $+\sec x \tan x$, and $\csc x$ gives $-\csc x \cot x$. Inside each pair, the co-function's derivative is the non-co-function's derivative with the sign flipped and every function swapped for its co-partner. So once you know that the derivative of $\sec x$ is $\sec x \tan x$, you get the cosecant formula by attaching a minus sign and turning $\sec$ into $\csc$ and $\tan$ into $\cot$. That is why memorizing three formulas plus the sign rule is enough for all six. NOTE (Memory rule): If the function name starts with c-o (cosine, cotangent, cosecant), its derivative is negative. Sine, tangent, and secant stay positive. This one rule fixes every sign in the table. ### Where the table comes from (Topic 2.10) Topic 2.10 does not ask you to memorize four more formulas from thin air. It asks you to rewrite tangent, cotangent, secant, and cosecant using sine and cosine, then differentiate with the quotient rule. That is the derivation, and it is worth doing once so the table stops feeling arbitrary. Take tangent as the model case. $$\tan'(x) = \left(\frac{\sin x}{\cos x}\right)' = \frac{\cos x \cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$$ The Pythagorean identity $\cos^2 x + \sin^2 x = 1$ collapses the numerator to $1$, so the result is $\dfrac{1}{\cos^2 x} = \sec^2 x$. The same rewrite-then-quotient-rule path produces the other three: cotangent from $\dfrac{\cos x}{\sin x}$, secant from $\dfrac{1}{\cos x}$, and cosecant from $\dfrac{1}{\sin x}$. Worked example 1 runs the secant case in full. The practical payoff is that you never have to gamble on a half-remembered formula: if you can write the function over sine and cosine, you can regenerate its derivative. ### Inverse trig derivatives (Topic 3.4) Topic 3.4 adds six more derivatives, and these are on both the AB and BC exams, not BC only. The inverse trig functions ($\sin^{-1} x$, also written $\arcsin x$, and its relatives) differentiate to algebraic expressions with no trig function left in the answer. That is the recognition cue: an inverse trig derivative turns a trig-flavored input into a plain rational or radical output. | Function | Derivative | | --- | --- | | $\sin^{-1} x$ | $\dfrac{1}{\sqrt{1-x^2}}$ | | $\cos^{-1} x$ | $-\dfrac{1}{\sqrt{1-x^2}}$ | | $\tan^{-1} x$ | $\dfrac{1}{1+x^2}$ | | $\cot^{-1} x$ | $-\dfrac{1}{1+x^2}$ | | $\sec^{-1} x$ | $\dfrac{1}{|x|\sqrt{x^2-1}}$ | | $\csc^{-1} x$ | $-\dfrac{1}{|x|\sqrt{x^2-1}}$ | Two patterns make these manageable. First, the same co-function negative rule applies: $\cos^{-1} x$, $\cot^{-1} x$, and $\csc^{-1} x$ are just the negatives of their partners, so you memorize only three forms. Second, there are only three distinct shapes: $\dfrac{1}{\sqrt{1-x^2}}$ for the sine and cosine inverses, $\dfrac{1}{1+x^2}$ for the tangent and cotangent inverses, and $\dfrac{1}{|x|\sqrt{x^2-1}}$ for the secant and cosecant inverses. The arctangent form $\dfrac{1}{1+x^2}$ shows up most on the exam, often hidden inside an integral, because reversing it is how you integrate $\dfrac{1}{1+x^2}$. When the input to an inverse trig function is anything other than a bare $x$, the chain rule attaches the inner derivative on top. For $\left(\sin^{-1}(u)\right)'$ you write $\dfrac{u'}{\sqrt{1-u^2}}$, and worked example 3 shows this with $u = x^2$. Missing that inner derivative is the single most common inverse-trig error. ### Common mistakes and method selection Most points lost on trig derivatives come from a short list of predictable slips. On multiple choice, the distractor answers are usually built from exactly these errors, so an answer that matches one is a reason to recheck rather than move on. - Dropping the chain rule. The derivative of $\sin(3x)$ is $3\cos(3x)$, not $\cos(3x)$. Always multiply by the inner derivative when a function sits inside the trig function. - Confusing $\sec^2 x$ with $\sec x \tan x$. The first is the derivative of tangent; the second is the derivative of secant. They look alike and get swapped constantly. - Sign errors on co-functions. Forgetting the minus on $\cos x$, $\cot x$, $\csc x$, or their inverses. The c-o rule exists to catch exactly this. - Working in degrees. The formula $\sin'(x) = \cos x$ holds only when $x$ is in radians. AP Calculus is always in radians, so never switch your calculator to degree mode for a derivative. - Forgetting the absolute value in $\left(\sec^{-1} x\right)' = \dfrac{1}{|x|\sqrt{x^2-1}}$. The absolute value keeps the derivative positive on both branches of the domain. For method selection, build a two-part read. First name the row: which of the six functions (or six inverse functions) is it? Then scan for structure around it, a product, a quotient, or an inner function, and layer the product, quotient, or chain rule on top of the table entry. Topic 3.5 is entirely about this layering, choosing and combining derivative rules, and trig functions are where it shows up most often. ### Worked examples #### Rebuilding the secant derivative from the quotient rule Problem: If $f(x) = \sec x$, show that $f'(x) = \sec x \tan x$. 1. Rewrite secant in terms of cosine using the identity $\sec x = \dfrac{1}{\cos x}$. Now the quotient rule applies with numerator $1$ and denominator $\cos x$. 2. Set up the quotient rule $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$ with $u = 1$, so $u' = 0$, and $v = \cos x$, so $v' = -\sin x$. 3. Substitute the pieces: $\dfrac{(0)(\cos x) - (1)(-\sin x)}{\cos^2 x} = \dfrac{\sin x}{\cos^2 x}$. 4. Split the single fraction into a product: $\dfrac{\sin x}{\cos^2 x} = \dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x}$. 5. Recognize each factor: $\dfrac{1}{\cos x} = \sec x$ and $\dfrac{\sin x}{\cos x} = \tan x$, so the product is $\sec x \tan x$. Answer: $f'(x) = \sec x \tan x$ #### Product rule layered with the chain rule Problem: Differentiate $f(x) = x^2\tan(3x)$. 1. Read the outer structure: $x^2$ and $\tan(3x)$ are multiplied, so this is a product rule problem (Topic 2.8). Let $p(x) = x^2$ and $q(x) = \tan(3x)$. 2. Differentiate the first factor with the power rule: $p'(x) = 2x$. 3. Differentiate the second factor. It is $\tan$ of an inner function $3x$, so the chain rule gives the table entry $\sec^2 u$ times the inner derivative: $q'(x) = \sec^2(3x)\cdot 3 = 3\sec^2(3x)$. 4. Apply the product rule, first-prime times second plus first times second-prime: $f'(x) = (2x)\tan(3x) + (x^2)\big(3\sec^2(3x)\big)$. 5. Clean up: $f'(x) = 2x\tan(3x) + 3x^2\sec^2(3x)$. Nothing combines, so this is fully simplified. Answer: $f'(x) = 2x\tan(3x) + 3x^2\sec^2(3x)$ #### Inverse trig with an inner function (Topic 3.4) Problem: Differentiate $g(x) = \sin^{-1}(x^2)$. 1. Identify the outer function as inverse sine with an inner function $u = x^2$. The rule is $\left(\sin^{-1}(u)\right)' = \dfrac{u'}{\sqrt{1 - u^2}}$. 2. Find the inner derivative: $u = x^2$ gives $u' = 2x$. 3. Substitute $u = x^2$ and $u' = 2x$ into the formula: $g'(x) = \dfrac{2x}{\sqrt{1 - (x^2)^2}}$. 4. Simplify the exponent inside the radical: $(x^2)^2 = x^4$, so $g'(x) = \dfrac{2x}{\sqrt{1 - x^4}}$. Answer: $g'(x) = \dfrac{2x}{\sqrt{1 - x^4}}$ ### FAQ Q: What is the derivative of tan x? A: $\tan'(x) = \sec^2 x$. It comes from writing $\tan x = \dfrac{\sin x}{\cos x}$ and applying the quotient rule; the Pythagorean identity collapses the numerator to $1$, leaving $\dfrac{1}{\cos^2 x} = \sec^2 x$. Q: Why is the derivative of cos x negative? A: Because cosine is a co-function, and every co-function (cosine, cotangent, cosecant) differentiates to a negative. Geometrically, $\cos x$ is decreasing wherever $\sin x$ is positive, so its slope $-\sin x$ must carry the minus sign. Q: Are inverse trig derivatives on the AP Calculus AB exam? A: Yes. Differentiating inverse trigonometric functions is Topic 3.4, which is part of both AB and BC. You need $\left(\tan^{-1} x\right)' = \dfrac{1}{1+x^2}$ and its relatives for both exams. Q: Do I have to memorize all six trig derivatives? A: Effectively three, plus the co-function sign rule. Memorize sine, tangent, and secant; get cosine, cotangent, and cosecant by flipping the sign and swapping each function for its co-partner. If any slip, rebuild them from sine and cosine with the quotient rule. ## Derivative of ln x, e^x, and a^x: Answers and Proofs URL: https://www.calclearn.org/guides/derivatives-of-exponentials-and-logs Last updated: 2026-07-22 Answer first: The derivative of ln x is 1/x, the derivative of e^x is e^x itself, and the derivative of a^x is a^x times ln a (so d/dx of 2^x is 2^x ln 2). Each is proved below: e^x from the limit defining e, ln x by inverting x = e^y, and a^x by rewriting a^x as e^(x ln a). ### Recognize the function before you pick a rule Three rules hide behind functions that look nearly identical, and choosing the wrong one is the most common way to lose these points. Before you differentiate, ask two questions. First: is the variable in the exponent or in the base? Second: if the variable is in the exponent, is the base $e$ or some other constant $a$? Your answers pick the rule. If the variable sits in the base with a constant exponent, like $x^5$, that is the power rule from Topic 2.5, not an exponential rule at all. The functions on this page put the variable somewhere else: in the exponent, as in $e^x$ and $a^x$, or inside a logarithm, as in $\ln x$. CED Topic 2.7 supplies the rules for $e^x$ and $\ln x$, and the $a^x$ result follows from a single rewrite that the later sections prove. The recognition habit worth building is to label the function by name (exponential, power, or logarithm) out loud before you write anything down, because the wrong label sends you to the wrong rule and no amount of careful algebra recovers from that. | Function | What to notice | Derivative | | --- | --- | --- | | $e^x$ | variable exponent, base $e$ | $e^x$ | | $a^x$ | variable exponent, constant base $a > 0$ | $a^x \ln a$ | | $\ln x$ | natural logarithm | $\frac{1}{x}$ | | $\log_a x$ | logarithm, base $a$ | $\frac{1}{x \ln a}$ | | $x^n$ | constant exponent (power rule, not here) | $n x^{n-1}$ | The $e^x$ and $\ln x$ rows are the Topic 2.7 rules to memorize; the $a^x$ and $\log_a x$ rows are not separate CED rules but derived results, following from those two through the $a^x = e^{x \ln a}$ and change-of-base rewrites proved in the later sections. NOTE (The a^x versus x^n trap): $2^x$ and $x^2$ are not the same kind of function. In $x^2$ the exponent is constant, so the power rule gives $2x$. In $2^x$ the exponent is the variable, so the exponential rule gives $2^x \ln 2$. Reading which slot the variable sits in, base or exponent, decides everything. Writing $\frac{d}{dx} 2^x = x \cdot 2^{x-1}$ applies the power rule to an exponential and is simply wrong. ### Why e^x is its own derivative Start from the limit definition of the derivative applied to $f(x) = e^x$. $$\frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h}$$ Use the exponent law $e^{x+h} = e^x \cdot e^h$, then factor $e^x$ out of the numerator. That factor does not depend on $h$, so it comes outside the limit. $$= \lim_{h \to 0} \frac{e^x(e^h - 1)}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h}$$ Everything now hinges on that last limit. The number $e$ is defined to be the one base for which $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$. For any other base the analogous limit is a different constant (it turns out to equal $\ln a$, which the $a^x$ section confirms). Because the base here is $e$, the limit is exactly $1$, and the derivative collapses to $e^x$ itself. $$\frac{d}{dx} e^x = e^x$$ That self-reproducing property, a function equal to its own rate of change, is why $e^x$ runs through exponential growth, differential equations, and Taylor series. Topic 2.7 lists it as a rule to memorize; the limit above is where the rule comes from. ### Proving d/dx ln x = 1/x with inverse functions (Topic 3.3) The natural logarithm is the inverse of $e^x$: $\ln x$ and $e^x$ undo each other. Topic 3.3 says you can differentiate an inverse function using the chain rule together with the defining relationship, and $\ln x$ is the cleanest example. Set $y = \ln x$. By the definition of the logarithm as an inverse, that is the same statement as this equation. $$e^y = x$$ Differentiate both sides with respect to $x$. The right side is just $x$, whose derivative is $1$. The left side is $e^y$, where $y$ is a function of $x$, so the chain rule applies: the derivative of $e^y$ is $e^y$ times $\frac{dy}{dx}$. $$e^y \cdot \frac{dy}{dx} = 1$$ Solve for $\frac{dy}{dx}$, then replace $e^y$ with $x$ using $e^y = x$ from the setup. $$\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$$ So $\frac{d}{dx} \ln x = \frac{1}{x}$, valid for $x > 0$, the domain of $\ln x$. A useful extension, $\frac{d}{dx} \ln|x| = \frac{1}{x}$ for all $x \ne 0$, covers negative inputs and shows up whenever you antidifferentiate $\frac{1}{x}$. This inverse-function move (write the inverse relation, differentiate implicitly, solve) is exactly the method Topic 3.3 asks you to reuse for inverse trig functions in Topic 3.4. ### a^x = e^(x ln a): one rewrite handles every base You do not need a separate rule for $a^x$; you need one algebra step. Any positive number $a$ can be written as $a = e^{\ln a}$, because $e$ and $\ln$ are inverses. Raise both sides to the $x$ power and use the exponent law $(e^b)^x = e^{bx}$. $$a^x = \left(e^{\ln a}\right)^x = e^{x \ln a}$$ Now $a^x$ is an exponential with base $e$ and inner function $x \ln a$, so the chain rule finishes it. Because $\ln a$ is a constant, the derivative of the inner function $x \ln a$ is just $\ln a$. $$\frac{d}{dx} a^x = \frac{d}{dx} e^{x \ln a} = e^{x \ln a} \cdot \ln a = a^x \ln a$$ This also settles the loose end from Section 2. Differentiating $a^x$ straight from the definition produces the factor $\lim_{h \to 0} \frac{a^h - 1}{h}$, and the proof above shows that factor equals $\ln a$. For base $e$, $\ln e = 1$, which is precisely why $e^x$ carries no extra constant. Logarithms with other bases yield to the same trick through the change-of-base identity $\log_a x = \frac{\ln x}{\ln a}$. Since $\frac{1}{\ln a}$ is a constant multiplier, differentiate using $\frac{d}{dx}\ln x = \frac{1}{x}$. $$\frac{d}{dx} \log_a x = \frac{1}{\ln a} \cdot \frac{d}{dx} \ln x = \frac{1}{x \ln a}$$ ### Where students lose points Almost every error on these derivatives is one of a handful of predictable slips. - Treating $a^x$ like a power. $\frac{d}{dx} 2^x$ is $2^x \ln 2$, not $x \cdot 2^{x-1}$. The exponent is the variable, so the power rule does not apply. - Dropping the $\ln a$ factor. $\frac{d}{dx} 5^x = 5^x \ln 5$; writing just $5^x$ forgets the constant that every base other than $e$ carries. - Forgetting the chain rule inside a logarithm. $\frac{d}{dx} \ln(3x^2 + 1)$ is $\frac{1}{3x^2 + 1} \cdot 6x$, not $\frac{1}{3x^2 + 1}$. The $\frac{1}{x}$ rule is really $\frac{1}{\text{inside}}$ times the derivative of the inside. - Mixing up the two directions. The derivative of $\ln x$ is $\frac{1}{x}$, but the antiderivative of $\frac{1}{x}$ is $\ln|x| + C$; keep which way you are going straight. - Assuming $e^{2x}$ differentiates to $e^{2x}$. Only $e^x$ is its own derivative. $\frac{d}{dx} e^{2x} = 2e^{2x}$, because the chain rule contributes the inner derivative $2$. NOTE (One-line check): After differentiating any exponential, ask whether the base was $e$. If not, a $\ln(\text{base})$ factor must appear. After differentiating any logarithm, ask whether the inside was more complicated than $x$; if so, a chain-rule factor must appear. ### Worked examples #### Separate exponential, power, and natural-exponential terms Problem: Differentiate $f(x) = 3^x + x^3 + e^x$. 1. Classify each term before touching a rule. $3^x$ is exponential (variable in the exponent, base $3$), $x^3$ is a power (variable in the base, constant exponent), and $e^x$ is the natural exponential. Three look-alike terms, three different rules. 2. Differentiate $3^x$ with the exponential rule: the derivative of $a^x$ is $a^x \ln a$. Here $a = 3$, giving $3^x \ln 3$. 3. Differentiate $x^3$ with the power rule: the derivative of $x^n$ is $n x^{n-1}$. Here $n = 3$, giving $3x^2$. 4. Differentiate $e^x$: it is its own derivative, $e^x$. 5. Add the three results in order: $f'(x) = 3^x \ln 3 + 3x^2 + e^x$. Answer: $f'(x) = 3^x \ln 3 + 3x^2 + e^x$ #### A logarithm that needs the chain rule Problem: Differentiate $g(x) = \ln(x^2 + 4)$. 1. Recognize the structure: a natural log wrapped around $x^2 + 4$. The inside is more complicated than $x$, so the $\frac{1}{x}$ rule needs a chain-rule factor. Name the inner function $u = x^2 + 4$. 2. Apply the chain rule: if $y = \ln u$, then $y' = \frac{1}{u} \cdot u'$. The outer log gives $\frac{1}{x^2 + 4}$, with the inside left intact. 3. Differentiate the inside: $u' = 2x$. 4. Multiply the outer and inner pieces: $g'(x) = \frac{1}{x^2 + 4} \cdot 2x$. 5. Combine into a single fraction: $g'(x) = \frac{2x}{x^2 + 4}$. Answer: $g'(x) = \frac{2x}{x^2 + 4}$ #### Read a limit as a known derivative (Topic 2.7) Problem: Evaluate $\lim_{h \to 0} \frac{2^h - 1}{h}$. 1. This looks like an indeterminate $\frac{0}{0}$ limit, but Topic 2.7 flags a shortcut: sometimes a limit is exactly the definition of a derivative you already know. Compare it to the derivative definition $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$. 2. Match the pieces. If $f(x) = 2^x$ and $a = 0$, then $f(a+h) = 2^{0+h} = 2^h$ and $f(a) = 2^0 = 1$, so $\frac{f(a+h) - f(a)}{h} = \frac{2^h - 1}{h}$. The limit is therefore $f'(0)$ for $f(x) = 2^x$. 3. Differentiate with the exponential rule: $f'(x) = 2^x \ln 2$. 4. Evaluate at $a = 0$: $f'(0) = 2^0 \ln 2 = 1 \cdot \ln 2 = \ln 2$. Answer: $\lim_{h \to 0} \frac{2^h - 1}{h} = \ln 2$ ### FAQ Q: What is the derivative of 2^x? A: $\frac{d}{dx} 2^x = 2^x \ln 2$. It follows the general exponential rule $\frac{d}{dx} a^x = a^x \ln a$ with $a = 2$. A frequent mistake is writing $x \cdot 2^{x-1}$, but that is the power rule, which applies only when the variable is in the base, not the exponent. Q: Why does a^x differentiate to a^x ln a instead of just a^x? A: Only base $e$ gives a derivative with no extra factor, because $e$ is defined so that $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$. Rewriting $a^x = e^{x \ln a}$ and applying the chain rule produces the inner derivative $\ln a$, so $\frac{d}{dx} a^x = a^x \ln a$. When $a = e$, $\ln e = 1$ and the factor disappears. Q: What is the derivative of ln(x)? A: $\frac{d}{dx} \ln x = \frac{1}{x}$ for $x > 0$. If the logarithm wraps a more complicated expression, say $\ln(x^2 + 4)$, the chain rule adds the derivative of the inside: $\frac{1}{x^2 + 4} \cdot 2x$. For negative inputs, $\frac{d}{dx} \ln|x| = \frac{1}{x}$ holds for all $x \ne 0$. ## Taylor and Maclaurin Series: How to Build Them URL: https://www.calclearn.org/guides/taylor-maclaurin-series Last updated: 2026-07-22 Answer first: A Taylor series centered at x = a sums f^(n)(a)/n! times (x - a)^n over all n from 0 up. A Maclaurin series is that series centered at 0. In practice you rarely compute every derivative: start from a known series like e^x, sin x, cos x, or 1/(1-x) and substitute, multiply, differentiate, or integrate instead. ### The Taylor series formula A Taylor series is what a Taylor polynomial (Topic 10.11) becomes when you never stop adding terms. You pick a center $x = a$, compute the derivatives of $f$ there, divide the $n$-th derivative by $n!$, and attach the result to $(x-a)^n$. Summing over every $n$ from $0$ up produces the series. $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots$$ Dividing by $n!$ is not arbitrary. It is exactly the factor that forces the polynomial's $n$-th derivative at $a$ to match $f^{(n)}(a)$, so each new term corrects the next derivative. That is the design principle behind the whole formula. The center $a$ is the point the series is built around and where its approximation is sharpest. When $a = 0$ the $(x-a)^n$ factors become plain powers of $x$, and the series gets its own name: the Maclaurin series. Maclaurin is not a separate idea, just Taylor centered at zero. $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)\,x + \frac{f''(0)}{2!}x^2 + \cdots$$ Every partial sum of a Taylor series is a Taylor polynomial (LIM-8.E.1), so the series is the limit of its polynomials. This is BC Topic 10.14, resting directly on the coefficient $\frac{f^{(n)}(a)}{n!}$ from Topic 10.11. All of Unit 10 is BC-only. ### Two ways to build a series, and how to choose The derivative formula defines a Taylor series, but on the exam it is usually the slow way in. Differentiating a function four or five times to spot a pattern is error-prone, and for a composite like $e^{-x^2}$ the derivatives grow fast: $f'(x) = -2x\,e^{-x^2}$, then $f''(x) = (4x^2 - 2)e^{-x^2}$, and it only gets messier. Finding a clean formula for $f^{(n)}(0)$ that way is brutal. The fast route is almost always to start from a series you already know and reshape it, so the first decision is recognition. Look at $f(x)$ and ask whether it is a close relative of one of the five standard series. If $f$ is a standard function with $x$ replaced by some expression (like $\cos(2x)$, $e^{-x^2}$, or $\frac{1}{1+x}$), multiplied by a power of $x$, or the derivative or integral of a standard function, build it from the known series. Only when the function is genuinely unlike anything on the list, or when the problem asks for a series centered at a nonzero $a$, should you fall back on computing derivatives one at a time. Example 3 is exactly that case: $\ln x$ centered at $x = 1$. NOTE (Recognition rule): Is the function a known series in disguise (a substitution, a product with $x^k$, or a derivative or integral of one)? Reshape the known series. Is it truly new, or centered at $a \ne 0$? Use the derivative formula. This is the whole game in Topics 10.14 and 10.15. The CED names $\sin x$, $\cos x$, and $e^x$ as the foundation for constructing other series (LIM-8.F.2) and $\frac{1}{1-x}$ as the geometric building block (LIM-8.F.1). Memorizing a handful of series buys you dozens more for almost no work. ### The five series worth memorizing These five carry the overwhelming majority of series questions on the BC exam. The first three are the CED-designated foundation, the fourth is a geometric series, and the fifth comes from substituting $-x$ into the fourth to get $\frac{1}{1+x}$, then integrating that term by term. | Function | Maclaurin series | Converges for | | --- | --- | --- | | $e^x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \cdots$ | all $x$ | | $\sin x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \cdots$ | all $x$ | | $\cos x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \cdots$ | all $x$ | | $\dfrac{1}{1-x}$ | $\displaystyle\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots$ | $|x| < 1$ | | $\ln(1+x)$ | $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} = x - \frac{x^2}{2} + \cdots$ | $-1 < x \le 1$ | The three that converge for every $x$ ($e^x$, $\sin x$, $\cos x$) are the workhorses, because you can substitute any expression into them without worrying about an interval. The trig pair also has a memorable structure: $\sin x$ uses only odd powers, because sine is an odd function, while $\cos x$ uses only even powers, because cosine is even, and both alternate in sign. If you can write the $e^x$ series, you can reconstruct $\sin x$ and $\cos x$ as its odd and even parts with alternating signs. ### Building from a known series Once you have a known series, four operations generate new ones, and all four are fair game (Topic 10.15). 1. Substitution: replace $x$ with any expression. Putting $-x^2$ into $e^x$ gives the series for $e^{-x^2}$; putting $2x$ into $\cos x$ gives the series for $\cos(2x)$. 2. Multiplication by a power of $x$: to get the series for $x^2 e^x$, multiply every term of the $e^x$ series by $x^2$. 3. Term-by-term differentiation: differentiating $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ gives $\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n\,x^{n-1}$. 4. Term-by-term integration: integrating $\frac{1}{1+x^2}$ term by term recovers the series for $\arctan x$. You can chain these freely. To build the series for $x^2 \cos(2x)$, substitute $2x$ into the cosine series, then multiply every term by $x^2$. Differentiation and integration do not change the radius of convergence (LIM-8.D.6), though the endpoint behavior can shift, so an interval-of-convergence question still needs both endpoints checked on their own (Topic 10.13). ### Common mistakes - Forgetting to divide by $n!$. The coefficient is $\frac{f^{(n)}(a)}{n!}$, not $f^{(n)}(a)$. This is the single most common slip. - Losing the $(x-a)$ shift. For a series centered at $a \ne 0$, the powers are $(x-a)^n$, not $x^n$. Writing $x$ alone silently recenters the series at $0$. - Recomputing derivatives when a substitution would do. If the function is a known series in disguise, differentiating from scratch wastes time and invites errors. - Mishandling signs after substitution. $(-x^2)^n = (-1)^n x^{2n}$: the exponent acts on both the sign and the variable. - Assuming the interval carries over unchanged. Substituting $u = g(x)$ into a series valid for $|u| < 1$ makes the new condition $|g(x)| < 1$, which can be a very different set of $x$-values. ### Worked examples #### Building $e^x$ from the derivative formula Problem: Find the Maclaurin series for $f(x) = e^x$ directly from the coefficient formula. 1. Every derivative of $e^x$ is again $e^x$, so $f'(x) = e^x$, $f''(x) = e^x$, and in general $f^{(n)}(x) = e^x$ for every $n$. 2. Evaluate at the center $x = 0$. Since $e^0 = 1$, every derivative gives the same value: $f^{(n)}(0) = 1$ for all $n$. 3. Form the coefficient $\frac{f^{(n)}(0)}{n!} = \frac{1}{n!}$, so the term of degree $n$ is $\frac{x^n}{n!}$. 4. Sum over all $n$: $e^x = \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ 5. A ratio test on $\frac{x^n}{n!}$ gives $L = 0$ for every fixed $x$, so the series converges for all real $x$. This is the anchor series: the other four in the table are variations on it or on the geometric series. Answer: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, converging for every real $x$. #### $e^{-x^2}$ by substitution, not brute force Problem: Find the Maclaurin series for $e^{-x^2}$. 1. Differentiating $e^{-x^2}$ repeatedly gets ugly fast, so recognize it instead: it is $e^u$ with $u = -x^2$. Start from the known series $e^u = \sum_{n=0}^{\infty} \frac{u^n}{n!}$. 2. Substitute $u = -x^2$ into that series: $e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}$. 3. Simplify the power: $(-x^2)^n = (-1)^n x^{2n}$, so $e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$. 4. Write out the first terms: $e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$ 5. Because the $e^u$ series converges for all $u$, the substituted series converges for all real $x$. Building it this way took three lines; the derivative formula would have taken a page. Answer: $e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2!} - \cdots$, valid for all real $x$. #### A Taylor series centered at a nonzero point Problem: Find the Taylor series for $f(x) = \ln x$ centered at $x = 1$. 1. There is no simpler known series sitting at $x = 1$, so use the derivative formula. Set $f(x) = \ln x$ with center $a = 1$; then $f(1) = \ln 1 = 0$, so the constant term vanishes. 2. Differentiate repeatedly: $f'(x) = x^{-1}$, $f''(x) = -x^{-2}$, $f'''(x) = 2x^{-3}$, $f^{(4)}(x) = -6x^{-4}$. 3. Evaluate at $x = 1$: $f'(1) = 1$, $f''(1) = -1$, $f'''(1) = 2$, $f^{(4)}(1) = -6$. The pattern is $f^{(n)}(1) = (-1)^{n-1}(n-1)!$ for $n \ge 1$. 4. Divide by $n!$ to get the coefficient, using $\frac{(n-1)!}{n!} = \frac{1}{n}$: $\frac{f^{(n)}(1)}{n!} = \frac{(-1)^{n-1}(n-1)!}{n!} = \frac{(-1)^{n-1}}{n}$. 5. Attach each coefficient to $(x-1)^n$: $\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots$ 6. Substituting $t = x - 1$ turns this into $\ln(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \cdots$, the standard $\ln(1+x)$ series from the table. That confirms the work and shows the two are the same series, just shifted. It converges for $-1 < t \le 1$, that is $0 < x \le 2$. Answer: $\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n$, converging for $0 < x \le 2$. ### FAQ Q: What is the difference between a Taylor series and a Maclaurin series? A: A Maclaurin series is a Taylor series centered at $x = 0$. That is the only difference. The general Taylor series is $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$; setting the center $a = 0$ turns the $(x-a)^n$ factors into $x^n$ and gives the Maclaurin series. Every Maclaurin series is a Taylor series, but a Taylor series centered at $a \ne 0$ is not Maclaurin. Q: Which Taylor series do I need to memorize for AP Calculus BC? A: At minimum, memorize the Maclaurin series for $e^x$, $\sin x$, $\cos x$, and $\frac{1}{1-x}$. The CED points to $\sin x$, $\cos x$, and $e^x$ as the foundation for building others (LIM-8.F.2), and $\frac{1}{1-x}$ is the geometric building block (LIM-8.F.1). Knowing $\ln(1+x)$ helps too, but you can recover it by integrating the series for $\frac{1}{1+x}$. Q: How do you find a Taylor series centered at a point other than zero? A: Use the derivative formula directly: compute $f(a), f'(a), f''(a), \dots$ at the center $a$, form the coefficients $\frac{f^{(n)}(a)}{n!}$, and attach each to $(x-a)^n$. The powers must be $(x-a)^n$, not $x^n$. Example 3 above builds $\ln x$ at $a = 1$ this way. Substituting a known series works only if you can rewrite $f$ in terms of $(x-a)$, such as writing $\ln x = \ln(1 + (x-1))$ and using the $\ln(1+t)$ series with $t = x - 1$. ## Separable Differential Equations: Step-by-Step Method URL: https://www.calclearn.org/guides/separable-differential-equations Last updated: 2026-07-22 Answer first: To solve a separable differential equation, get every y term with dy on one side and every x term with dx on the other, integrate both sides, add a single +C, then solve for y. If an initial condition is given, substitute it to find C right after integrating, then isolate y and state any domain restriction. ### Recognizing a separable equation A differential equation is separable when you can rearrange it so that every $y$ sits with $dy$ on one side and every $x$ sits with $dx$ on the other. Formally, the right side of the equation has to factor into a function of $x$ alone times a function of $y$ alone: $$\frac{dy}{dx} = g(x) \cdot h(y)$$ Recognition is the entire skill in Topic 7.6, and it comes down to one question: can I split the right side into an $x$-only piece and a $y$-only piece? Multiplication and division separate cleanly, so a product or quotient of $x$ and $y$ terms always works. Addition and subtraction do not, so $\frac{dy}{dx} = x + y$ cannot be separated, and neither can anything where $x$ and $y$ are locked inside the same function, like $\frac{dy}{dx} = \sin(xy)$. On the AP exam, a first-order equation you are asked to solve algebraically will always separate. | Equation | Separable? | Why | | --- | --- | --- | | $\frac{dy}{dx} = 6xy$ | Yes | $(6x)(y)$: an $x$-piece times a $y$-piece | | $\frac{dy}{dx} = \frac{x^2}{y}$ | Yes | $(x^2)\left(\frac{1}{y}\right)$ | | $\frac{dy}{dx} = x + y$ | No | a sum will not factor into $x$ times $y$ | | $\frac{dy}{dx} = \sin(xy)$ | No | $x$ and $y$ trapped inside one function | Run this factoring test before you write anything else. It decides whether the rest of the method even applies, and if it fails, the exam wants a slope field or Euler's method instead of algebra. ### The method, step by step Once an equation passes the separable test, the procedure is mechanical and identical every time. 1. Separate. Multiply and divide until all the $y$ terms (including $dy$) are on one side and all the $x$ terms (including $dx$) are on the other. 2. Integrate both sides. Antidifferentiate each side with respect to its own variable. 3. Add one constant. Put a single $+C$ on the $x$ side. You do not need a constant on both sides, because two arbitrary constants collapse into one. 4. Apply the initial condition (if the problem gives one) and solve for $C$ now, before you isolate $y$. 5. Solve for $y$. Write $y$ as an explicit function of $x$, and state any domain restriction. The move in step 1, treating $dy$ and $dx$ as quantities you can multiply across, looks like a shortcut but it is really bookkeeping for the chain rule. Integrating $p(y) \cdot \frac{dy}{dx}$ with respect to $x$ is the same as integrating $p(y)$ with respect to $y$, which is exactly why the two sides can be antidifferentiated in their own variables. NOTE (One constant, not two): Writing $+C$ on both sides is not wrong, just wasted work: subtracting leaves a single constant $C = C_2 - C_1$. Put one $+C$ on the $x$ side and move on. The general solution of a first-order equation carries exactly one arbitrary constant. ### Initial conditions and particular solutions A general solution like $y = Ae^{3x^2}$ describes infinitely many curves, one for each value of the constant. An initial condition, a single point the solution must pass through, selects exactly one of them: the particular solution (Topic 7.7). This is the difference between a family of curves and the one curve the problem is actually about. Apply the initial condition right after you integrate, while $C$ is still alone on one side. Substitute the point, solve for $C$, then isolate $y$. Solving for $y$ first and substituting afterward gives the same answer, but it drags the constant through more algebra and is where sign errors creep in. If you have time, verify the result the way Topic 7.2 asks: differentiate your answer, confirm it satisfies the original equation, and check that it passes through the given point. That catches most algebra slips before they cost you. ### Domain restrictions: the point students lose This is the step that separates a full-credit free-response answer from one that drops a point. A particular solution is a single continuous, differentiable function through the initial point, so its domain is the largest interval that contains that point and on which the solution stays defined. The answer is not finished until you state that interval. Two things cut the interval short. A vertical asymptote is the first: if the solution blows up at some $x = a$, the particular solution lives only on the side of $a$ that holds the initial condition. The original equation is the second: if $\frac{dy}{dx}$ divides by $y$, the solution can never reach a point where $y = 0$, so those points bound the interval. Report the domain explicitly as an interval and confirm the initial $x$-value sits inside it (FUN-7.E.3). ### Mistakes that cost points A handful of errors show up again and again on graded work. - Adding the constant too late. Put $+C$ in at the integration step and carry it through, not after you exponentiate. - Losing the absolute value. $\int \frac{1}{y}\,dy = \ln|y|$, not $\ln y$. The absolute value is what lets the final constant be positive or negative. - Dropping the constant solution. Dividing by $y$ assumes $y \neq 0$, but $y = 0$ (or whatever makes $h(y) = 0$) is often a solution too. It usually reappears when you let the constant equal zero. - Choosing the wrong sign. When solving for $y$ produces a $\pm$, the initial condition decides which sign to keep. - Skipping the domain. If the problem gives an initial condition, the interval of validity is part of the answer, not an afterthought. ### Where this fits in Unit 7 Separation of variables is the only algebraic solving method the course expects you to run by hand, which makes recognizing it the real decision. If the equation separates, use this method. If it does not, the exam turns to qualitative tools: a slope field to picture the family of solutions (Topics 7.3 and 7.4), or Euler's method to approximate a value numerically (Topic 7.5, BC only). Two famous models are separable equations in disguise. Exponential growth and decay, $\frac{dy}{dt} = ky$, separates to give $y = y_0 e^{kt}$ (Topic 7.8). The BC logistic equation $\frac{dy}{dt} = ky(a - y)$ separates too, though it needs partial fractions to integrate (Topic 7.9). Both are just this same method applied to a specific right side. ### Worked examples #### Finding a general solution Problem: Find the general solution of $\frac{dy}{dx} = 6xy$. 1. Check that it is separable. The right side is $(6x)(y)$, an $x$-only factor times a $y$-only factor, so separation of variables applies (Topic 7.6). 2. Separate the variables. Divide both sides by $y$ and multiply by $dx$ to get $\frac{1}{y}\,dy = 6x\,dx$. Dividing by $y$ assumes $y \neq 0$; keep that case in mind. 3. Integrate both sides. $\int \frac{1}{y}\,dy = \int 6x\,dx$ gives $\ln|y| = 3x^2 + C$, with a single constant on the right. 4. Solve for $y$. Exponentiate both sides: $|y| = e^{3x^2 + C} = e^C \cdot e^{3x^2}$. Drop the absolute value by letting the constant absorb the sign, $A = \pm e^C$, so $y = Ae^{3x^2}$. 5. Recover the lost solution. The excluded case $y = 0$ is itself a solution of the original equation, and it matches $y = Ae^{3x^2}$ when $A = 0$. Letting $A$ be any real number captures every solution. Answer: $y = Ae^{3x^2}$, where $A$ is any constant. #### Particular solution with a domain restriction Problem: Solve $\frac{dy}{dx} = y^2$ with the initial condition $y(0) = 1$, and state the domain of the solution. 1. Confirm separability. $y^2 = (1)(y^2)$ is a function of $x$ (the constant $1$) times a function of $y$, so the equation separates. 2. Separate. $\frac{1}{y^2}\,dy = dx$, which is $y^{-2}\,dy = dx$. 3. Integrate both sides. $\int y^{-2}\,dy = \int 1\,dx$ gives $-\frac{1}{y} = x + C$. 4. Apply the initial condition now, while $C$ is alone. Substitute $x = 0$ and $y = 1$: $-\frac{1}{1} = 0 + C$, so $C = -1$. 5. Solve for $y$. The equation becomes $-\frac{1}{y} = x - 1$, so $\frac{1}{y} = 1 - x$ and $y = \frac{1}{1 - x}$. 6. State the domain. The solution has a vertical asymptote at $x = 1$. The initial point $x = 0$ lies to the left of it, so the particular solution is valid only on $x < 1$. Answer: $y = \frac{1}{1 - x}$ for $x < 1$. #### Choosing the correct sign from the initial condition Problem: Solve $\frac{dy}{dx} = -\frac{x}{y}$ with $y(0) = 2$. 1. Separate. Multiply both sides by $y$ and by $dx$: $y\,dy = -x\,dx$. The equation divides by $y$, so $y = 0$ is excluded from the start. 2. Integrate both sides. $\int y\,dy = \int -x\,dx$ gives $\frac{y^2}{2} = -\frac{x^2}{2} + C$. 3. Clear fractions. Multiply through by $2$ and rename the constant: $y^2 = -x^2 + C_1$, so $x^2 + y^2 = C_1$. 4. Apply the initial condition. Substitute $x = 0$ and $y = 2$: $0 + 4 = C_1$, so $x^2 + y^2 = 4$. 5. Solve for $y$ and pick the sign. Taking the square root gives $y = \pm\sqrt{4 - x^2}$. Because $y(0) = 2$ is positive, keep the positive root: $y = \sqrt{4 - x^2}$. 6. State the domain. The expression needs $4 - x^2 > 0$, and the original equation forbids $y = 0$, which occurs at $x = \pm 2$. Both conditions give $-2 < x < 2$. Answer: $y = \sqrt{4 - x^2}$ for $-2 < x < 2$. ### FAQ Q: Are separable differential equations on the AB exam or only BC? A: Both. Topics 7.6 and 7.7 are part of the AB and BC curriculum, and separation of variables is a common free-response task on each exam. Only Euler's method (7.5) and logistic models (7.9) in this unit are BC-only. Q: Do I need a $+C$ on both sides when I integrate? A: No. Integrating both sides technically produces two constants, but their difference is a single arbitrary constant, so you write one $+C$, usually on the $x$ side. A first-order equation has exactly one constant in its general solution. Q: Should I solve for y before or after using the initial condition? A: Use the initial condition first, right after integrating, while the constant is alone on one side. Solve for $C$, then isolate $y$. That order keeps the algebra short and avoids sign mistakes. Q: Why does the AP rubric care about the domain of the solution? A: A particular solution has to be one continuous function through the initial point, so it cannot cross a vertical asymptote or a point where the equation is undefined. Stating the correct interval (FUN-7.E.3) is a scored part of the answer. ## Slope Fields and Euler's Method Explained URL: https://www.calclearn.org/guides/slope-fields-eulers-method Last updated: 2026-07-22 Answer first: A slope field draws a tiny segment at every grid point whose slope equals dy/dx there, so you see the shape of solution curves without solving the equation. Euler's method (BC only, Topic 7.5) approximates one value by stepping along that field: new y = old y + (step size)(slope). ### What a slope field shows You are looking at a slope field problem when a differential equation is written in the form $\frac{dy}{dx} = f(x, y)$ and the task is to sketch it, match it to a graph, or describe its solution curves, not to find a formula. A slope field (FUN-7.C.1) is a grid of short line segments. At each point the segment is drawn with the slope that the differential equation assigns to that point. Because $\frac{dy}{dx}$ is the slope of the solution curve passing through a point, the field is a picture of every solution at once. $$\frac{dy}{dx} = f(x, y)$$ This is the method-selection idea behind Topics 7.3 and 7.4: a slope field gives qualitative information (where solutions rise, fall, flatten, or steepen) with no integration required. You never solve the equation to draw the field. You only evaluate $f(x, y)$ at grid points. That makes slope fields the right tool when the equation is hard or impossible to solve by hand but you still need to understand how its solutions behave. Unit 7 is worth 5 to 10 percent of both the AB and BC exams, and this qualitative reasoning shows up in both multiple-choice and free-response. ### Sketching a slope field (Topic 7.3) The procedure is pure evaluation. There is no antiderivative anywhere in it. 1. Read the slope rule $f(x, y)$ straight off the right side of the differential equation. 2. Take each grid point $(x, y)$ and substitute to get the numerical slope there. 3. Draw a short segment at that point with that slope: flat when the slope is $0$, tilted up to the right when positive, down when negative, and steeper as the magnitude grows. 4. Repeat across the whole grid. Segments that share a slope reveal the shape of the solutions. A fast shortcut is the isocline: set $f(x, y)$ equal to a constant to find every point sharing one slope. Setting $f(x, y) = 0$ locates all the horizontal segments, where solution curves have horizontal tangents: the candidate locations for a peak or a valley (you must still check that $\frac{dy}{dx}$ changes sign, or the sign of $\frac{d^2y}{dx^2}$, to confirm an extremum). For $\frac{dy}{dx} = x - y$, the zero-slope isocline is the line $y = x$, so every solution levels off the moment it crosses that line. Example 1 builds this field point by point. ### Reasoning with slope fields on the exam Topic 7.4 turns the picture into reasoning, and three question types come up again and again. Matching an equation to a field: test one or two easy points instead of the whole grid. If $f$ depends only on $x$, every segment in the same vertical column is identical; if $f$ depends only on $y$, every segment in the same horizontal row is identical. That single observation eliminates most wrong answers instantly. Sketching a particular solution: start at the given initial-condition point and draw a curve that stays tangent to the nearby segments, following them like a current (FUN-7.C.3). Your curve hugs the segments. It never crosses through one. As a quick matching example, suppose one candidate is $\frac{dy}{dx} = x$ and another is $\frac{dy}{dx} = y$. The first depends only on $x$, so its segments are flat all along the $y$-axis and grow steeper to the left and right; the second depends only on $y$, so its segments are flat all along the $x$-axis and grow steeper up and down. Reading which axis carries the horizontal segments settles the match in seconds. Confirming concavity is the third type and the one students lose points on. Differentiate $\frac{dy}{dx}$ again with respect to $x$, treating $y$ as a function of $x$, to get $\frac{d^2y}{dx^2}$. Its sign tells you where the solution is concave up or down. This links Unit 7 back to the derivative reasoning of Unit 5 and is a standard free-response step. Example 2 works it in full. ### Euler's method: stepping through the field (BC only) NOTE (BC only): Euler's method is Topic 7.5 and appears on the BC exam only (FUN-7.C.4). AB students can stop after the slope field sections above. A slope field shows the shape of solutions; Euler's method puts a number on one of them. It approximates the value of a particular solution at a target $x$ by taking short straight steps along the slope field. Start at the initial condition, follow the current slope for a small step of width $h$, land on a new point, read the new slope there, and repeat until you reach the target. $$y_{n+1} = y_n + h \cdot f(x_n, y_n), \qquad x_{n+1} = x_n + h$$ Organize the work in a table. Each row records the current point, the slope $f(x_n, y_n)$ there, the step change $\Delta y = h \cdot f(x_n, y_n)$, and the resulting next value of $y$. The table below runs Euler's method for $\frac{dy}{dx} = x + y$ with $y(0) = 1$ and $h = 0.5$, which is worked step by step in Example 3. | Step $n$ | $x_n$ | $y_n$ | Slope $x_n + y_n$ | $\Delta y = h \cdot \text{slope}$ | $y_{n+1}$ | | --- | --- | --- | --- | --- | --- | | 0 | 0 | 1 | 1 | 0.5 | 1.5 | | 1 | 0.5 | 1.5 | 2 | 1 | 2.5 | A smaller step size $h$ means more rows and less error. Because every step follows a straight tangent line, the estimate drifts to one side of the true curve in a predictable way: when the solution is concave up, the tangent steps sit below it and Euler underestimates; when it is concave down, the steps sit above and Euler overestimates. Decide which case you are in from the sign of $\frac{d^2y}{dx^2}$, never by guessing. ### Common mistakes - Solving the differential equation when the problem only asks for a sketch. A slope field needs evaluation, not integration. - Drawing curved marks. Every field segment is straight; only the solution curve you sketch through them is allowed to bend. - In Euler's method, forgetting to multiply the slope by $h$. The step change is always $\Delta y = h \cdot f(x_n, y_n)$, not just $f(x_n, y_n)$. - Reusing the starting slope for every step. Recompute $f(x_n, y_n)$ at each new point before taking the next step. - Guessing whether the estimate is an overestimate or underestimate. Read it off the concavity of the solution instead. ### Worked examples #### Sketching the slope field for dy/dx = x - y Problem: Compute the slopes of $\frac{dy}{dx} = x - y$ at the nine grid points with $x = -1, 0, 1$ and $y = -1, 0, 1$, then describe the resulting slope field. 1. The slope rule is $f(x, y) = x - y$. At each grid point, subtract the $y$-coordinate from the $x$-coordinate to get the slope of the segment drawn there. 2. Bottom row, $y = -1$: at $(-1, -1)$ the slope is $-1 - (-1) = 0$; at $(0, -1)$ it is $0 - (-1) = 1$; at $(1, -1)$ it is $1 - (-1) = 2$. 3. Middle row, $y = 0$: at $(-1, 0)$ the slope is $-1 - 0 = -1$; at $(0, 0)$ it is $0$; at $(1, 0)$ it is $1$. 4. Top row, $y = 1$: at $(-1, 1)$ the slope is $-1 - 1 = -2$; at $(0, 1)$ it is $0 - 1 = -1$; at $(1, 1)$ it is $1 - 1 = 0$. 5. Draw a short segment at each point with that slope: flat where the slope is $0$, tilted up to the right where it is positive, down where it is negative, and steeper as the magnitude grows toward $\pm 2$. 6. Notice the flat segments at $(-1, -1)$, $(0, 0)$, and $(1, 1)$. They all lie on the line $y = x$, which is the zero-slope isocline found by setting $x - y = 0$. Along that line every solution curve momentarily levels off. Answer: The slope at each point equals $x - y$, ranging from $-2$ at the top left to $2$ at the bottom right, with horizontal segments all along the line $y = x$. #### Concavity of a solution from the equation Problem: For the same equation $\frac{dy}{dx} = x - y$, the particular solution passes through $(0, 2)$. Is that solution increasing or decreasing there, is it concave up or concave down, and where are its tangent lines horizontal? 1. Increasing versus decreasing is set by $\frac{dy}{dx}$. At $(0, 2)$: $\frac{dy}{dx} = 0 - 2 = -2$. The slope is negative, so the solution is decreasing at that point. 2. Concavity comes from $\frac{d^2y}{dx^2}$. Differentiate both sides of $\frac{dy}{dx} = x - y$ with respect to $x$, treating $y$ as a function of $x$, so that $y$ differentiates to $\frac{dy}{dx}$: this gives $\frac{d^2y}{dx^2} = 1 - \frac{dy}{dx}$. 3. Substitute the original equation $\frac{dy}{dx} = x - y$ to write the second derivative in terms of $x$ and $y$: $\frac{d^2y}{dx^2} = 1 - (x - y) = 1 - x + y$. 4. Evaluate at $(0, 2)$: $\frac{d^2y}{dx^2} = 1 - 0 + 2 = 3$. This is positive, so the solution is concave up at $(0, 2)$: it is decreasing while bending back upward. 5. Horizontal tangents occur where $\frac{dy}{dx} = 0$, that is where $x - y = 0$, or $y = x$. The solution through $(0, 2)$ has a horizontal tangent at the moment it reaches the line $y = x$, exactly the isocline from Example 1. Answer: At $(0, 2)$ the solution is decreasing (slope $-2$) and concave up (second derivative $3$); its tangent line is horizontal wherever the solution crosses the line $y = x$. #### Euler's method table (BC only) Problem: Use Euler's method with step size $h = 0.5$ to approximate $y(1)$ for $\frac{dy}{dx} = x + y$ with the initial condition $y(0) = 1$. 1. Set up the recursion $y_{n+1} = y_n + h \cdot f(x_n, y_n)$ with $f(x, y) = x + y$ and $h = 0.5$. Reaching $x = 1$ from $x = 0$ in steps of $0.5$ takes two steps. 2. Step 1 from $(0, 1)$. The slope is $f(0, 1) = 0 + 1 = 1$. The change is $\Delta y = h \cdot \text{slope} = 0.5 \cdot 1 = 0.5$. So $y_1 = 1 + 0.5 = 1.5$ at $x_1 = 0.5$. 3. Step 2 from $(0.5, 1.5)$. Recompute the slope at the new point: $f(0.5, 1.5) = 0.5 + 1.5 = 2$. The change is $\Delta y = 0.5 \cdot 2 = 1$. So $y_2 = 1.5 + 1 = 2.5$ at $x_2 = 1$. 4. The two steps give $y(1) \approx 2.5$, matching the table in the Euler's method section. As a check, the exact solution is $y = 2e^x - x - 1$, so $y(1) = 2e - 2 \approx 3.44$. Euler undershoots because this solution is concave up: $\frac{d^2y}{dx^2} = \frac{dy}{dx} + 1 = x + y + 1 > 0$ along the path, so the straight tangent steps ride below the true curve. Answer: $y(1) \approx 2.5$. Because the solution is concave up, this is an underestimate of the true value $2e - 2 \approx 3.44$. ### FAQ Q: Is Euler's method on the AP Calculus AB exam? A: No. Euler's method is Topic 7.5 and is tested on the BC exam only. Slope fields (Topics 7.3 and 7.4) are on both AB and BC. Q: Does Euler's method overestimate or underestimate? A: It depends on concavity. When the solution curve is concave up, the tangent steps stay below it and Euler underestimates; when it is concave down, Euler overestimates. Check the sign of $\frac{d^2y}{dx^2}$ to decide rather than guessing. Q: Do I have to solve the differential equation to sketch its slope field? A: No, and that is the whole point. You only evaluate $\frac{dy}{dx} = f(x, y)$ at each grid point and draw a segment with that slope. Slope fields work even when the equation cannot be solved by hand. Q: What is the difference between a slope field and Euler's method? A: A slope field is a qualitative picture of every solution at once; Euler's method is a numerical procedure that estimates one solution's value at a specific $x$. The field shows shape, Euler produces a number. ## Parametric and Polar Calculus: Derivatives and Area URL: https://www.calclearn.org/guides/parametric-polar-calculus Last updated: 2026-07-22 Answer first: Which formula to use depends on what you are given and what is asked: divide dy/dt by dx/dt for a parametric slope, integrate the speed sqrt((dx/dt)^2+(dy/dt)^2) for arc length, or take one-half the integral of r^2 dtheta for polar area. The skill is reading the prompt to pick which one applies. ### Recognize which Unit 9 tool the problem wants Unit 9 is BC only and worth 11 to 12 percent of the exam. It bundles three curve systems that run on one engine, the chain rule. Parametric equations give $x$ and $y$ separately as functions of a parameter $t$. Polar equations give a radius $r$ as a function of an angle $\theta$. Vector-valued functions repackage the same parametric information as a position vector. Because the three setups look different on the page, the first move on any Unit 9 problem is not computation. It is matching what you are given, and what is asked, to the right formula. Read every prompt for two signals: what you are handed, and what it wants. If you have $x(t)$ and $y(t)$ and the question asks for a slope, reach for the parametric derivative. If that same curve is asked for the length of its path, reach for the arc length integral instead. A polar equation $r=f(\theta)$ paired with the word area almost always means the one-half r-squared integral, while a request for a tangent line to a polar curve means you rewrite it in $x$ and $y$ first. The table below is the decision map for the entire unit. The payoff for slowing down at this step is large. A student who writes the polar area integral when the question asked for a slope has lost the points before touching any algebra. Method selection, not execution, is where most of the exam's Unit 9 difficulty actually lives, and it is the one habit that carries across all nine topics. | What you are given | What is asked | Formula to reach for | | --- | --- | --- | | $x(t)$ and $y(t)$ | Slope of the tangent | $\frac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ (9.1) | | $x(t)$ and $y(t)$ | Concavity | $\dfrac{d^2y}{dx^2}$ (9.2) | | $x(t)$ and $y(t)$ | Length of the path | $\int_a^b\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt$ (9.3) | | Velocity vector, time interval | Total distance | $\int \text{speed}\,dt$ (9.6) | | $r=f(\theta)$ | Slope of the tangent | Rewrite as $x,y$, then $\frac{dy}{dx}$ (9.7) | | $r=f(\theta)$ | Area swept once | $\frac{1}{2}\int r^2\,d\theta$ (9.8) | | Two polar curves | Area between them | $\frac{1}{2}\int (r_{\text{outer}}^2-r_{\text{inner}}^2)\,d\theta$ (9.9) | ### Parametric derivatives: slope first, then concavity For a curve defined by $x(t)$ and $y(t)$, the slope $\frac{dy}{dx}$ is the ratio of the two rates of change (Topic 9.1). Divide the vertical rate by the horizontal rate. $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\qquad \frac{dx}{dt}\neq 0$$ The formula holds wherever $\frac{dx}{dt}$ is not zero. Where $\frac{dx}{dt}=0$ but $\frac{dy}{dt}\neq 0$, the tangent line is vertical and the slope is undefined, which is often exactly what a free-response part is testing. Where both rates are zero at once, the point may be a cusp, and you have to look closer. The second derivative (Topic 9.2) costs students more points than any other step in the unit, because the reflex is to differentiate $\frac{dy}{dx}$ with respect to $x$. You cannot, since everything is written in $t$. Differentiate the slope with respect to $t$, then divide by $\frac{dx}{dt}$ one more time. $$\frac{d^2y}{dx^2}=\frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$$ NOTE (The denominator trap): The bottom of the second-derivative formula is $\frac{dx}{dt}$, never $\frac{d^2x}{dt^2}$. Once you have $\frac{d^2y}{dx^2}$, its sign reports concavity exactly as it does for an ordinary function: positive means concave up, negative means concave down. ### Arc length and motion in the plane Arc length (Topic 9.3) measures the distance traveled along a curve as $t$ runs across an interval $[a,b]$. Square the two rates, add them, take the root, and integrate. $$L=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$$ The integrand is a speed, so a well-designed no-calculator problem almost always hides a perfect square you can pull out from under the radical. When the radical refuses to simplify, the problem is calculator-active: set the integral up correctly and evaluate it numerically. Getting the setup right earns the setup points even if the arithmetic runs long. Topics 9.4 through 9.6 restate all of this in vector language. A particle's position is $\langle x(t), y(t)\rangle$. Differentiate component by component for the velocity vector $\left\langle \frac{dx}{dt}, \frac{dy}{dt}\right\rangle$, and differentiate again for acceleration. Speed is the magnitude of velocity, which is the same radical sitting inside the arc length integral. $$\text{speed}=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},\qquad \text{distance}=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$$ That link is worth committing to memory. Total distance traveled is the integral of speed, and it equals the arc length of the path. Displacement is a different quantity: it is the integral of the velocity vector taken component by component, and it can be far smaller than distance whenever the particle reverses direction. Mixing up the two is a reliable way to lose an otherwise easy free-response point. ### Polar derivatives and area A polar curve $r=f(\theta)$ is secretly parametric, with $\theta$ as the parameter, through $x=r\cos\theta$ and $y=r\sin\theta$ (Topic 9.7). To find the slope of a tangent line, differentiate both with the product rule and divide, just as in the parametric case. $$\frac{dy}{dx}=\frac{\dfrac{dr}{d\theta}\sin\theta+r\cos\theta}{\dfrac{dr}{d\theta}\cos\theta-r\sin\theta}$$ Horizontal tangents occur where the numerator $\frac{dy}{d\theta}=0$ while the denominator is nonzero; vertical tangents occur where the denominator $\frac{dx}{d\theta}=0$ while the numerator $\frac{dy}{d\theta}\neq 0$. Setting $\frac{dr}{d\theta}=0$ and calling the result a horizontal tangent is a common and costly mistake, because $r$ is only one piece of $y=r\sin\theta$. Polar area (Topic 9.8) sums thin circular sectors instead of rectangles, which is why the formula carries a factor of one-half and squares the radius. $$A=\frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta=\frac{1}{2}\int_{\alpha}^{\beta}\bigl[f(\theta)\bigr]^2\,d\theta$$ The limits $\alpha$ and $\beta$ are the angles that sweep the region exactly once. For a region between two polar curves (Topic 9.9), subtract the inner radius squared from the outer radius squared inside a single integral. $$A=\frac{1}{2}\int_{\alpha}^{\beta}\left(r_{\text{outer}}^2-r_{\text{inner}}^2\right)\,d\theta$$ The difficulty here is entirely in the setup. Find where the curves meet by solving $f(\theta)=g(\theta)$, decide which curve is the outer boundary on each interval, and pick limits that trace the region a single time. Sketching both curves before you integrate is not optional; it is how you catch a region that has to be split into two integrals, and how you notice a curve that passes through the pole (the origin). Symmetry is your best ally for choosing limits. Many exam curves (cardioids, rose petals, limacons) are symmetric about an axis, so you can integrate over half the region and double the result, which keeps the arithmetic short and lowers the chance of a bookkeeping slip. For a rose like $r=\cos(3\theta)$, one petal runs between two consecutive angles where $r=0$, and multiplying a single petal's area by the number of petals is far safer than trying to sweep the whole flower in one integral. ### Where BC students lose points Most Unit 9 errors are setup errors, not algebra errors. These are the ones graders see most often, and each one is avoidable by reading the prompt before reaching for a formula. Slow down for one deliberate beat to name the technique, then execute; that single habit recovers more Unit 9 points than any amount of extra computational speed. - Dividing by $\frac{d^2x}{dt^2}$ instead of $\frac{dx}{dt}$ when computing the parametric second derivative. - Dropping the factor of one-half, or forgetting to square the radius, in the polar area formula. - Trying to integrate $\cos^2\theta$ or $\sin^2\theta$ directly instead of using the power-reduction identity first. - Using $\frac{dr}{d\theta}=0$ to locate horizontal tangents on a polar curve instead of $\frac{dy}{d\theta}=0$. - Choosing polar limits that trace a region twice, which doubles the area; sketch first and confirm that $\theta$ sweeps the region once. - Confusing displacement (the integral of the velocity vector) with total distance (the integral of speed). ### Worked examples #### Parametric slope and concavity at a point Problem: A curve is defined by $x=t^2$ and $y=t^3-3t$. Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ at $t=2$, and write the equation of the tangent line there. 1. Differentiate each equation with respect to $t$: $\frac{dx}{dt}=2t$ and $\frac{dy}{dt}=3t^2-3$. 2. Form the slope as the ratio of these rates (Topic 9.1): $\frac{dy}{dx}=\frac{3t^2-3}{2t}$, valid where $\frac{dx}{dt}=2t\neq 0$. 3. Evaluate at $t=2$: $\frac{dy}{dx}=\frac{3(4)-3}{2(2)}=\frac{9}{4}$. 4. For the second derivative (Topic 9.2), first rewrite the slope as $\frac{dy}{dx}=\frac{3}{2}\left(t-t^{-1}\right)$, then differentiate with respect to $t$: $\frac{d}{dt}\!\left(\frac{dy}{dx}\right)=\frac{3}{2}\left(1+t^{-2}\right)$. 5. Divide that by $\frac{dx}{dt}=2t$, not by $\frac{d^2x}{dt^2}$: $\frac{d^2y}{dx^2}=\frac{\frac{3}{2}\left(1+t^{-2}\right)}{2t}=\frac{3\left(t^2+1\right)}{4t^3}$. 6. At $t=2$: $\frac{d^2y}{dx^2}=\frac{3(4+1)}{4(8)}=\frac{15}{32}>0$, so the curve is concave up there. 7. Find the point of tangency: $x=2^2=4$ and $y=2^3-3(2)=2$. The tangent line is $y-2=\frac{9}{4}(x-4)$. Answer: dy/dx = 9/4 and d^2y/dx^2 = 15/32 at t = 2. The tangent line is y - 2 = (9/4)(x - 4), and the curve is concave up there. #### Arc length of a parametric curve Problem: Find the length of the curve $x=\frac{1}{3}t^3$, $y=\frac{1}{2}t^2$ for $0\le t\le 2$. 1. Read the cue: a parametric curve with a $t$-interval and the word length points to the arc length integral (Topic 9.3), not a slope or an area formula. 2. Differentiate: $\frac{dx}{dt}=t^2$ and $\frac{dy}{dt}=t$. 3. Build the integrand $\sqrt{(dx/dt)^2+(dy/dt)^2}=\sqrt{t^4+t^2}$. 4. Factor $t^2$ out of the radical. Since $t\ge 0$ on this interval, $\sqrt{t^2\left(t^2+1\right)}=t\sqrt{t^2+1}$. 5. Set up the definite integral: $L=\int_0^2 t\sqrt{t^2+1}\,dt$. 6. Substitute $u=t^2+1$, so $du=2t\,dt$ and $t\,dt=\frac{1}{2}\,du$. When $t=0$, $u=1$; when $t=2$, $u=5$. 7. Integrate: $L=\frac{1}{2}\int_1^5 \sqrt{u}\,du=\frac{1}{2}\cdot\frac{2}{3}u^{3/2}\Big|_1^5=\frac{1}{3}\left(5^{3/2}-1\right)$. 8. Simplify $5^{3/2}=5\sqrt{5}$: $L=\frac{1}{3}\left(5\sqrt{5}-1\right)\approx 3.39$. Answer: L = (1/3)(5*sqrt(5) - 1), which is about 3.39. #### Area enclosed by a cardioid Problem: Find the area enclosed by the polar curve $r=1+\cos\theta$. 1. A single polar curve traced once, plus the word area, points to $A=\frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$ (Topic 9.8). The cardioid is traced exactly once as $\theta$ runs from $0$ to $2\pi$. 2. Set up the integral: $A=\frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2\,d\theta$. 3. Expand the square: $(1+\cos\theta)^2=1+2\cos\theta+\cos^2\theta$. 4. Replace $\cos^2\theta$ with the power-reduction identity $\cos^2\theta=\frac{1+\cos 2\theta}{2}$, the step students skip most often: $A=\frac{1}{2}\int_0^{2\pi}\left(1+2\cos\theta+\frac{1+\cos 2\theta}{2}\right)\,d\theta$. 5. Combine the constants, $1+\frac{1}{2}=\frac{3}{2}$: $A=\frac{1}{2}\int_0^{2\pi}\left(\frac{3}{2}+2\cos\theta+\frac{1}{2}\cos 2\theta\right)\,d\theta$. 6. Integrate term by term. Over a full period, $\int_0^{2\pi} 2\cos\theta\,d\theta=0$ and $\int_0^{2\pi}\frac{1}{2}\cos 2\theta\,d\theta=0$, while $\int_0^{2\pi}\frac{3}{2}\,d\theta=3\pi$. 7. So $A=\frac{1}{2}(3\pi)=\frac{3\pi}{2}$. Answer: A = 3*pi/2. ### FAQ Q: Is parametric and polar calculus on the AP Calculus AB exam? A: No. All of Unit 9 is BC only. AB students are not tested on parametric derivatives, polar area, or vector-valued functions. On the BC exam the unit carries 11 to 12 percent of the score. Q: How do you find the second derivative of a parametric equation? A: Find $\frac{dy}{dx}$ first, then differentiate that expression with respect to $t$ and divide by $\frac{dx}{dt}$. In symbols, $\frac{d^2y}{dx^2}=\dfrac{\frac{d}{dt}(dy/dx)}{dx/dt}$. Dividing by $\frac{d^2x}{dt^2}$ is the classic wrong turn. Q: Why is there a one-half in the polar area formula? A: Because polar area adds up thin circular sectors, not rectangles. A sector of radius $r$ and angle $d\theta$ has area $\frac{1}{2}r^2\,d\theta$, and integrating those sectors gives $A=\frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$. Q: How do you find horizontal tangents on a polar curve? A: Convert to $y=r\sin\theta$, then set $\frac{dy}{d\theta}=0$ and check that $\frac{dx}{d\theta}\neq 0$ at that angle. Do not set $\frac{dr}{d\theta}=0$; that finds where the radius stops changing, not where the tangent is horizontal. ## Continuity and the Three Types of Discontinuity URL: https://www.calclearn.org/guides/continuity-and-discontinuities Last updated: 2026-07-22 Answer first: A function is continuous at x = c when f(c) exists, the limit as x approaches c exists, and the two are equal. If any one fails you get a discontinuity: removable (a hole where the limit exists but misses f(c)), jump (left and right limits differ), or infinite (a vertical asymptote). ### The three-part continuity test A function $f$ is continuous at $x = c$ when three conditions all hold at once: $f(c)$ is defined, $\lim_{x \to c} f(x)$ exists, and those two values are equal (Topic 1.11). Check them in that order. If $f(c)$ is undefined there is nothing for the limit to match; if the limit does not exist there is no single value to match; and even when both exist, they can disagree. A discontinuity is just the name for the first condition that fails, and which condition fails, together with what the one-sided limits are doing, tells you the type. $$f(c)\text{ is defined},\qquad \lim_{x \to c} f(x)\text{ exists},\qquad \lim_{x \to c} f(x) = f(c)$$ Graphically, continuity at $c$ means you can trace the graph straight through $x = c$ without lifting your pencil. A function is continuous on an interval when it is continuous at every point of that interval (Topic 1.12); at an endpoint you only need the appropriate one-sided limit. The standard function families, polynomial, rational, power, exponential, logarithmic, and trigonometric, are continuous at every point of their domains, so on the exam the only places worth checking are where a denominator equals zero or where a piecewise rule switches from one formula to another. ### The three types of discontinuity Every discontinuity the AB and BC exams ask you to classify is one of three kinds (Topic 1.10), and the quickest way to tell them apart is to ask what the two one-sided limits are doing. | Type | One-sided limits | Overall limit | Graph | | --- | --- | --- | --- | | Removable | Both equal the same finite number | Exists, but $\neq f(c)$ or $f(c)$ undefined | A single hole | | Jump | Both finite but unequal | Does not exist | A break, the graph steps | | Infinite | At least one is $+\infty$ or $-\infty$ | Does not exist | A vertical asymptote | The removable discontinuity is the only one whose limit exists, and that single fact is worth memorizing. It is why this type, and only this type, can be repaired: the graph already agrees on the value it approaches from both sides, it just fails to actually take that value at $c$. Jump and infinite discontinuities have no limit at all, so no single value can patch them. In terms of the three-part test: a removable discontinuity fails only the last condition (or $f(c)$ is simply undefined), a jump fails the second condition because the one-sided limits disagree, and an infinite discontinuity also fails the second condition, with the function running off to infinity near $c$. ### How to spot which discontinuity you have Recognizing which discontinuity you are looking at, before grinding through arithmetic, is the skill the exam is really testing. Two setups cover almost everything you will see. Rational functions: factor the numerator and denominator completely, then examine each factor that makes the denominator zero. If that same factor also appears in the numerator and cancels completely, the discontinuity there is removable, a hole in the graph. If the factor remains in the denominator after canceling, the function grows without bound near that point and you have an infinite discontinuity, that is, a vertical asymptote. A single rational expression essentially never produces a jump. Piecewise functions: the only points that can fail are the boundaries where the definition changes formula. At each boundary, compute the one-sided limit from the piece on each side. If the two one-sided limits agree with each other and with the function value there, $f$ is continuous. If both are finite but unequal, it is a jump. If either piece is unbounded, it is infinite. Jumps almost always come from piecewise definitions, so this is where to expect them. ### Removing a discontinuity When a limit exists at a discontinuity, you can remove it by redefining the function at that one point so it equals the limit (Topic 1.13). If $\lim_{x \to c} f(x) = L$, then declaring $f(c) = L$ fills the hole and makes $f$ continuous at $c$. This only works for removable discontinuities: jump and infinite discontinuities have no limit, so changing a single value cannot rescue them. The same principle powers the common 'find the constant' problems. To make a piecewise function continuous at a boundary, set the expression from each side and the function value equal at that point, then solve for the unknown constant. Worked Example 2 runs this all the way through. ### The Intermediate Value Theorem The Intermediate Value Theorem is an existence theorem: it promises that a value is achieved somewhere without telling you where (Topic 1.16). If $f$ is continuous on the closed interval $[a, b]$ and $d$ is any number between $f(a)$ and $f(b)$, then there is at least one $c$ in $(a, b)$ with $f(c) = d$. $$f\text{ continuous on }[a,b],\ \ d\text{ between }f(a)\text{ and }f(b)\ \implies\ f(c)=d\ \text{ for some }c\in(a,b)$$ Most IVT questions ask you to show that an equation has a solution. The routine is fixed: name a closed interval, state that $f$ is continuous on it, evaluate $f$ at both endpoints, and show the target value $d$ is caught between the two outputs, usually because $f$ changes sign so $d = 0$ sits between a negative and a positive value. Never skip the continuity step. A function with a jump can leap straight over $d$ without ever equaling it, so the theorem simply does not apply without continuity on the entire closed interval. ### Common mistakes A few traps show up again and again. - A hole is still a discontinuity. A limit existing does not make a function continuous; you also need $f(c)$ to exist and to equal that limit. - A vertical asymptote means the limit does not exist. Infinity is not a real number, so writing that a limit 'equals infinity' is only shorthand for the limit failing to exist. - The IVT gives you at least one $c$, not exactly one, and it never tells you the value of $c$. - The IVT needs continuity on the whole closed interval $[a, b]$. Confirming the endpoints alone is not enough. ### Worked examples #### Classify every discontinuity of a rational function Problem: Find and classify all discontinuities of $f(x) = \dfrac{x^2 - 4}{x^2 - x - 2}$. 1. Factor the numerator and denominator. The numerator is $x^2 - 4 = (x-2)(x+2)$ and the denominator is $x^2 - x - 2 = (x-2)(x+1)$. 2. The denominator is zero at $x = 2$ and $x = -1$, so those are the only places $f$ can be discontinuous. Everywhere else $f$ is a rational function on its domain, so it is continuous. 3. Cancel the common factor: $f(x) = \dfrac{(x-2)(x+2)}{(x-2)(x+1)} = \dfrac{x+2}{x+1}$ for $x \neq 2$. The factor $(x-2)$ cancels completely, so the discontinuity at $x = 2$ is removable. 4. Find the limit at the hole: $\lim_{x \to 2} f(x) = \dfrac{2+2}{2+1} = \dfrac{4}{3}$. There is a hole at $\left(2, \tfrac{4}{3}\right)$. 5. At $x = -1$ the factor $(x+1)$ stays in the denominator. As $x \to -1$ the numerator approaches $1$ while the denominator approaches $0$, so $f$ is unbounded. That is an infinite discontinuity, a vertical asymptote at $x = -1$. Answer: Removable discontinuity (hole) at x = 2 with limit 4/3, and an infinite discontinuity (vertical asymptote) at x = -1. #### Make a piecewise function continuous Problem: Find the value of $c$ that makes $f$ continuous everywhere, where $f(x) = \begin{cases} cx + 1 & x \le 2 \\ x^2 - c & x > 2 \end{cases}$ 1. Each piece is a polynomial, so the only place continuity can fail is the boundary $x = 2$. Apply the three-part test there. 2. Function value: the first piece defines $f(2)$, so $f(2) = c \cdot 2 + 1 = 2c + 1$. 3. Left-hand limit: for $x \to 2^-$ use the first piece, so $\lim_{x \to 2^-} f(x) = 2c + 1$. 4. Right-hand limit: for $x \to 2^+$ use the second piece, so $\lim_{x \to 2^+} f(x) = 2^2 - c = 4 - c$. 5. For the overall limit to exist, the one-sided limits must be equal: $2c + 1 = 4 - c$. Solve: $3c = 3$, so $c = 1$. 6. Check: with $c = 1$, the left side gives $2(1)+1 = 3$, the right side gives $4 - 1 = 3$, and $f(2) = 3$. All three agree, so $f$ is continuous. For any other value of $c$ the one-sided limits differ and $x = 2$ is a jump discontinuity. Answer: c = 1. Any other value leaves a jump discontinuity at x = 2. #### Use the IVT to show a root exists Problem: Show that $f(x) = x^3 + x - 1$ has at least one real root in the interval $[0, 1]$. 1. Check the hypothesis first. $f$ is a polynomial, so it is continuous on the closed interval $[0, 1]$, and the IVT applies. 2. Evaluate at the endpoints: $f(0) = 0^3 + 0 - 1 = -1$ and $f(1) = 1^3 + 1 - 1 = 1$. 3. Identify the target value. A root means $f(c) = 0$, so $d = 0$. Since $f(0) = -1 < 0 < 1 = f(1)$, the value $0$ lies between $f(0)$ and $f(1)$. 4. Apply the IVT: because $f$ is continuous on $[0, 1]$ and $0$ is between $f(0)$ and $f(1)$, there is at least one $c$ in $(0, 1)$ with $f(c) = 0$. That $c$ is the root. Answer: Because f is continuous on [0, 1] with f(0) = -1 and f(1) = 1, the IVT guarantees at least one c in (0, 1) where f(c) = 0. ### FAQ Q: Is a removable discontinuity still a discontinuity if the limit exists? A: Yes. The limit existing satisfies only the second condition of continuity. The function is still discontinuous at that point because $f(c)$ is either undefined or not equal to the limit, so the graph has a hole. It becomes continuous only once you redefine $f(c)$ to equal that limit. Q: What is the difference between a jump and an infinite discontinuity? A: Both have no overall limit, but for different reasons. At a jump the left and right one-sided limits are both finite, they just do not match, so the graph steps up or down. At an infinite discontinuity the function is unbounded, at least one one-sided limit runs off to $+\infty$ or $-\infty$, which shows up as a vertical asymptote. Q: Does the Intermediate Value Theorem tell you where the root is? A: No. The IVT only guarantees that at least one such point exists somewhere in the open interval; it does not locate it and does not say how many there are. To find or estimate the actual value you need another method, such as solving the equation or a numerical technique. ## How to Study for AP Calculus AB: Unit-by-Unit Plan URL: https://www.calclearn.org/guides/how-to-study-ap-calculus-ab Last updated: 2026-07-22 Answer first: Study by exam weight, not chapter order. Units 5 and 6 are the heaviest, each 15-20% of the multiple-choice, so start there, then Units 1, 2, 4, and 8. Before computing anything, name the method a problem needs; that recognition step is what the exam rewards and what solvers cannot do for you. ### Read the exam before you study for it AP Calculus AB is passable with steady work. In May 2025, 64.2% of the 286,722 students who took it scored a 3 or higher, and 20.3% earned a 5, with a mean score of 3.21. Those odds tilt in your favor when your studying matches how the exam is built rather than how a textbook is ordered. The exam runs 3 hours 15 minutes and splits into two sections across four parts. Section I is 45 multiple-choice questions; Section II is 6 free-response questions. Each section has a no-calculator part and a calculator part, and that split should shape how you practice. | Section / part | Questions | Calculator | Time | Weight | | --- | --- | --- | --- | --- | | I, Part A (MC) | 30 | Not permitted | 60 min | 33.3% | | I, Part B (MC) | 15 | Required | 45 min | 16.7% | | II, Part A (FRQ) | 2 | Required | 30 min | 16.7% | | II, Part B (FRQ) | 4 | Not permitted | 60 min | 33.3% | NOTE (Practice the way the exam is weighted): The two no-calculator parts (Section I Part A and Section II Part B) are 66.7% of your score. Most of your practice should be pencil and paper, so the machine never becomes a crutch you cannot use on exam day. ### Weight your time by what the exam weights The College Board publishes a weighting band for each unit's share of the multiple-choice section. Two units, 5 and 6, sit at the top at 15-20% each. Spend the most time there, then on the 10-15% units, and give lighter targeted review to the 5-10% units. | Unit | Topic | AB weight | Study emphasis | | --- | --- | --- | --- | | 1 | Limits and Continuity | 10-15% | High | | 2 | Differentiation: Definition and Fundamental Properties | 10-15% | High | | 3 | Differentiation: Composite, Implicit, and Inverse Functions | 5-10% | Medium | | 4 | Contextual Applications of Differentiation | 10-15% | High | | 5 | Analytical Applications of Differentiation | 15-20% | Highest | | 6 | Integration and Accumulation of Change | 15-20% | Highest | | 7 | Differential Equations | 5-10% | Medium | | 8 | Applications of Integration | 10-15% | High | Read the arithmetic. Units 5 and 6 together can be as much as 40% of the multiple-choice weight. Add Units 1, 2, 4, and 8 and you have the large majority of the exam. Units 3 and 7 are lighter, but they are cheap points because their methods are narrow: master the handful of moves and you rarely miss them. ### Study each unit for its recognition trigger The goal for each unit is not just to execute a method but to recognize when a problem calls for it. Here is the trigger to master in each unit, with the CED topic numbers so you can drill the exact skill. - Unit 1 (10-15%): Given a limit, try direct substitution first. A number is the answer. A $\frac{0}{0}$ result is a signal that tells you the tool: factor and cancel for polynomials, multiply by the conjugate for radicals, combine fractions for compound quotients. Topic 1.7 (Selecting Procedures for Determining Limits) tests exactly this choice. - Unit 2 (10-15%): Know the derivative rules cold, including the derivatives of $\sin x$, $\cos x$, $e^x$, and $\ln x$ (Topic 2.7). The trigger is the shape of the function: a product needs the product rule, a quotient needs the quotient rule. - Unit 3 (5-10%): The chain rule is your response to a composition, a function nested inside another (Topic 3.1). Implicit differentiation is your response when $y$ is tangled with $x$ and cannot be isolated (Topic 3.2). Topic 3.5 asks you to select among the derivative procedures. - Unit 4 (10-15%): This unit is word problems. Two changing quantities linked by an equation, asked for a rate, is a related rate (Topics 4.4-4.5). A $\frac{0}{0}$ or $\frac{\infty}{\infty}$ limit that resists algebra is L'Hospital's Rule (Topic 4.7). Position, velocity, and acceleration language is straight-line motion (Topic 4.2). - Unit 5 (15-20%): The heaviest analytical unit. Increasing and decreasing and local extrema point to the first derivative; concavity and inflection points point to the second. Absolute max or min on a closed interval $[a, b]$ is always the Candidates Test (Topic 5.5). Maximize or minimize a real quantity is optimization (Topics 5.10-5.11). - Unit 6 (15-20%): The other heaviest unit, and the trigger is the integrand. A basic power or known form integrates directly; an integrand holding a function together with a constant multiple of its own derivative is u-substitution (Topic 6.9). Topic 6.14 is the selection skill itself. Note that integration by parts and partial fractions (Topics 6.11-6.12) are BC only, so AB antidifferentiation is substitution or algebraic rearrangement. - Unit 7 (5-10%): A differential equation $\frac{dy}{dx}$ that separates onto opposite sides is solved by separation of variables (Topics 7.6-7.7). A described growth or decay situation is an exponential model (Topic 7.8). Slope fields (Topics 7.3-7.4) are read, not solved. Euler's method and logistic models are BC only. - Unit 8 (10-15%): Integration applied. Area between curves integrates top minus bottom (Topics 8.4-8.6). Volume of a solid of revolution is the disk or washer method (Topics 8.9-8.12); a solid with known cross sections integrates the cross-sectional area (Topics 8.7-8.8). Average value of a function has its own formula (Topic 8.1). ### Method selection is the skill the exam actually tests The CED builds method selection into the course on purpose. Topics 1.7, 3.5, and 6.14 are each named Selecting Procedures or Selecting Techniques, and they introduce no new formulas. They test one thing: can you look at a problem and name the right tool before you compute? A solver will crank out any integral you hand it. The exam rewards the student who knows which integral this is and why. Turn that into a habit. For every practice problem, write one sentence before any algebra: this is a $\frac{0}{0}$ limit with a radical, so I use the conjugate, or this asks for absolute extrema on a closed interval, so I use the Candidates Test. Students who skip this step burn time on the no-calculator parts, where 66.7% of the score lives and there is no machine to fall back on. ### Where AB students lose points Most lost points on this exam are not hard concepts. They are habits. Drill these out before May. - Dropping the $+ C$ on an indefinite integral, or forgetting to change the limits of integration after a u-substitution in a definite integral (Topic 6.9). - Leaving out the differential, writing an integral with no $dx$. Free-response readers score notation, and Communication and Notation is assessed only on the free-response section. - Not justifying. Writing that $f$ has a local maximum at $x = 2$ earns nothing without the reason: because $f'$ changes from positive to negative there. The free-response section weights justification heavily. - Skipping units in applied answers. A rate problem answered as 12 instead of 12 liters per minute loses the point. - Reaching for the calculator by reflex. Practice Section I Part A and Section II Part B entirely by hand so the exam feels the same as your prep. ### Worked examples #### Choosing a Limit Method by Its Indeterminate Form Problem: Evaluate $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$. 1. Start with the default move: substitute $x = 4$. The numerator is $\sqrt{4} - 2 = 0$ and the denominator is $4 - 4 = 0$, so the limit has the indeterminate form $\frac{0}{0}$. A number would have been the answer; $\frac{0}{0}$ is a signal, not a result. 2. Read the form to pick the tool. The obstacle is a radical in the numerator, so multiply the numerator and denominator by the conjugate $\sqrt{x} + 2$. 3. Multiply: $\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{(\sqrt{x})^2 - 2^2}{(x - 4)(\sqrt{x} + 2)} = \frac{x - 4}{(x - 4)(\sqrt{x} + 2)}$. 4. Cancel the common factor $x - 4$, which is valid because $x \to 4$ means $x \neq 4$: the expression becomes $\frac{1}{\sqrt{x} + 2}$. 5. Now direct substitution works: $\frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$. Answer: The limit equals 1/4. The recognition step (0/0 with a radical means use the conjugate) is what the exam tests; the algebra is routine once you have chosen the method. #### Recognizing When an Integral Calls for u-Substitution Problem: Find $\int 2x\sqrt{x^2 + 1}\,dx$. 1. Scan the integrand for a function nested inside another. Here $x^2 + 1$ sits inside the square root. Check whether its derivative appears as a factor: $\frac{d}{dx}(x^2 + 1) = 2x$, and $2x$ is indeed multiplying the root. An inside function together with its derivative is the trigger for u-substitution (Topic 6.9). 2. Let $u = x^2 + 1$. Then $\frac{du}{dx} = 2x$, so $du = 2x\,dx$. 3. Rewrite the whole integral in terms of $u$. The $2x\,dx$ becomes $du$ and $\sqrt{x^2 + 1}$ becomes $\sqrt{u}$: $\int \sqrt{u}\,du = \int u^{1/2}\,du$. 4. Integrate with the power rule for antiderivatives, raising the exponent by one and dividing: $\int u^{1/2}\,du = \frac{u^{3/2}}{3/2} + C = \frac{2}{3}u^{3/2} + C$. 5. Substitute back $u = x^2 + 1$: $\frac{2}{3}(x^2 + 1)^{3/2} + C$. Answer: The antiderivative is (2/3)(x^2+1)^(3/2) + C. On AB, an integral like this is substitution or nothing: integration by parts and partial fractions are BC only, so if basic rules and a u-sub do not crack an AB integral, recheck your algebra. #### Using the Candidates Test for Absolute Extrema Problem: Find the absolute maximum and minimum values of $f(x) = x^3 - 3x^2 + 1$ on the closed interval $[-1, 3]$. 1. Recognize the phrasing. Absolute max and min on a closed interval names one method: the Candidates Test (Topic 5.5). By the Extreme Value Theorem, because $f$ is continuous on $[-1, 3]$ the extrema exist, and they can occur only at critical points or endpoints. 2. Find the critical points. Differentiate: $f'(x) = 3x^2 - 6x = 3x(x - 2)$. Set $f'(x) = 0$: $3x(x - 2) = 0$ gives $x = 0$ and $x = 2$. Both lie in $[-1, 3]$, and $f'$ exists everywhere, so these are the only critical points. 3. List the candidates: the two critical points $x = 0$ and $x = 2$, plus the two endpoints $x = -1$ and $x = 3$. 4. Evaluate $f$ at each. $f(-1) = (-1)^3 - 3(-1)^2 + 1 = -1 - 3 + 1 = -3$. $f(0) = 0 - 0 + 1 = 1$. $f(2) = 2^3 - 3(2)^2 + 1 = 8 - 12 + 1 = -3$. $f(3) = 3^3 - 3(3)^2 + 1 = 27 - 27 + 1 = 1$. 5. Compare the four outputs. The largest is $1$ (at $x = 0$ and $x = 3$); the smallest is $-3$ (at $x = -1$ and $x = 2$). Answer: The absolute maximum value is 1 and the absolute minimum value is -3. You never needed the first or second derivative test here: on a closed interval, comparing values at the candidates is faster, and that is exactly what the Candidates Test is for. ### FAQ Q: Is AP Calculus AB hard? A: It is manageable with steady practice. In 2025, 64.2% of students scored 3 or higher and 20.3% earned a 5, with a mean of 3.21. The students who struggle usually know the mechanics but freeze on choosing which method a problem needs, and that is a habit you can train. Q: How many hours should I study for AP Calculus AB? A: There is no fixed number, but a useful rule is to weight your time by exam weight. Put the most into Units 5 and 6 (each 15-20% of the multiple-choice), a solid block into Units 1, 2, 4, and 8, and lighter targeted review into Units 3 and 7. Steady weekly practice beats a cram. Q: What is the hardest unit in AP Calculus AB? A: By weight, Units 5 (analytical applications of derivatives) and 6 (integration) carry the most and are worth the most study. Unit 5 leans on justification and Unit 6 on choosing the right antidifferentiation technique, which is where a lot of points are won or lost. Q: Can I use a calculator on the AP Calculus AB exam? A: Only on part of it. A graphing calculator is required for Section I Part B and Section II Part A, and not permitted for Section I Part A or Section II Part B. The two no-calculator parts are 66.7% of the score, so practice most of your work by hand. ## AP Calculus AB vs BC: Difficulty, Units, Which to Take URL: https://www.calclearn.org/guides/ap-calculus-ab-vs-bc Last updated: 2026-07-22 Answer first: AP Calculus BC covers everything in AB plus two extra units (parametric/polar and infinite series) and a few added techniques, on an identically structured exam. BC is broader, not harder. Take BC if your precalc is strong and you want more college credit; take AB for a steadier pace. ### The short answer: BC is AB plus two units AP Calculus BC is not a harder remix of AB. It is AB plus more. Every topic in the AB course, from limits in Unit 1 through volumes of revolution in Unit 8, is also in BC. On top of that shared core, BC adds two full units and a short list of extra techniques inside the units both courses share. That is the whole difference. Picture AB as the first eight units of one book; BC is those same eight units plus chapters nine and ten. The exams are built the same way. Both run 3 hours and 10 minutes, both have 42 multiple-choice questions and 6 free-response questions, and both use the same calculator rules: no calculator on the larger multiple-choice part and the final free-response part, calculator required on the other two. BC even reports a separate AB subscore, so one BC exam tells a college how you did on the AB core and on the full BC course. Many high schools run BC as a single year that covers what colleges often split into Calculus I and Calculus II. ### What BC adds on top of AB Six extra topics live inside the units AB and BC share, and two entire units belong to BC alone. Here is the full list of what a BC student learns that an AB student never touches. | What BC adds | CED location | What it is | | --- | --- | --- | | Integration by parts | Topic 6.11 | Reverses the product rule for products like $x\cos x$ | | Linear partial fractions | Topic 6.12 | Splits a rational function into simpler fractions to integrate | | Improper integrals | Topic 6.13 | Integrals with an infinite limit or an unbounded integrand | | Euler's method | Topic 7.5 | Steps along a slope field to approximate a solution numerically | | Logistic models | Topic 7.9 | Growth that slows as it nears a carrying capacity | | Arc length | Topic 8.13 | Length of a smooth curve | | Parametric, polar, and vector-valued functions | Unit 9 (10-15%) | Motion and curves described without $y=f(x)$ | | Infinite sequences and series | Unit 10 (15-20%) | Convergence tests, power series, and Taylor and Maclaurin series | Units 9 and 10 together are 25 to 35 percent of the BC exam, and Unit 10, infinite sequences and series, is tied with Unit 6 (Integration and Accumulation of Change) as the largest single block (15 to 20 percent) and is the most genuinely new way of thinking in the course. Everything else on BC is AB content, often weighted a little lighter to make room. That reweighting is why the limits and basic derivatives worth 10 to 15 percent each on AB drop to 5 to 10 percent each on BC. ### What the 2025 score data really means Here are the numbers people quote. In 2025, 78.6 percent of BC students scored 3 or higher against 64.2 percent of AB students, and 43.9 percent of BC students earned a 5 against 20.3 percent on AB. The mean BC score was 3.82; the mean AB score was 3.21. Read quickly, that looks like proof BC is the easier exam. It is not. BC covers strictly more material on an identically structured test, so nothing about the exam itself is easier. The gap comes from who sits for each one. BC is self-selected. Students reach it after doing well in earlier math, usually on an accelerated track, and students who found precalculus a grind are steered toward AB instead. So the BC pass rate is measuring a stronger, pre-filtered group, not a gentler test. AB draws the larger and far more mixed population: 286,722 students took AB in 2025 versus 160,954 for BC. Put that same BC cohort in front of the AB exam and its scores would be higher still. NOTE (The AB subscore): Every BC exam also produces an AB subscore, a separate 1 to 5 score calculated from the portion of the test that covers AB-level content. A BC student who never quite mastered series can still earn a strong AB subscore, and the college credit that comes with it. That is a real safety net AB-only students do not have. ### How to choose: AB or BC The honest deciding factor is not raw ability, it is precalculus fluency and pace. BC moves through roughly a third more content in the same school year, so the week-to-week speed is the real jump, not the difficulty of any single idea. Use these signals. Lean toward BC if most of these fit: - Precalculus (algebra, trig identities, function manipulation) feels fast and reliable, not effortful. - You are on an accelerated math track or can commit to a faster pace all year. - Your target colleges award more credit for BC, often two semesters of calculus instead of one. - You are heading toward engineering, physics, math, or economics, where series and parametric motion show up early. Lean toward AB if most of these fit: - You want to build the core ideas of calculus at a steadier pace, with room to master each one. - Precalculus took real effort, and you would rather be solid than fast. - You plan to take BC or college Calculus II later; AB first, then BC, is a common and reasonable two-year route. Both choices are legitimate. A 5 on AB is a genuine achievement and full calculus credit at most schools. Taking AB first does not put you behind; it changes the pace, not the ceiling. ### Misconceptions that trip people up A few beliefs about the AB/BC choice are common and wrong. - BC is the hard class and AB is the easy one. BC is broader, not built from harder ideas. Series (Unit 10) is the one genuinely new mode of thinking; the rest is AB content at a faster clip. - A low BC score is worse than a high AB score for credit. Colleges set their own policies, so check each school. The BC exam's AB subscore often earns AB-equivalent credit even when the overall BC score is modest. - The BC exam is longer or has more questions. It is identical to AB in length, question count, and calculator rules. - You should always take AB before BC to be safe. It is a valid path, but strong students routinely take BC in one year and finish two semesters of college calculus. The extra year is insurance, not a requirement. - Precalculus does not matter much for BC. It matters more, because the faster pace leaves less time to shore up shaky algebra or trig along the way. ### Worked examples #### Shared AB and BC content: a u-substitution integral Problem: Evaluate $\int_0^1 2x(x^2+1)^3\,dx$. 1. Both AB and BC test this. Read the integrand: the inside function is $x^2+1$, and its derivative $2x$ is sitting right there as a factor. That derivative pairing is the u-substitution signal (Topic 6.9), a technique on both exams. 2. Let $u = x^2+1$. Then $du = 2x\,dx$, which is exactly the $2x\,dx$ already in the integral. 3. Convert the limits to $u$-values so you never switch back: when $x=0$, $u = 0+1 = 1$; when $x=1$, $u = 1+1 = 2$. 4. Rewrite the whole integral in $u$: $\int_0^1 2x(x^2+1)^3\,dx = \int_1^2 u^3\,du$. 5. Integrate and evaluate: $\int_1^2 u^3\,du = \frac{u^4}{4}\Big|_1^2 = \frac{2^4}{4} - \frac{1^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$. Answer: $\frac{15}{4}$ #### A BC-only technique: integration by parts Problem: Evaluate $\int x\cos x\,dx$. 1. Classify the integrand. It is a product of an algebraic factor $x$ and a trigonometric factor $\cos x$, and neither is the derivative of the other's inside, so u-substitution has nothing to cancel. That is the integration-by-parts signal (Topic 6.11), which appears only on BC. 2. Choose $u$ and $dv$ with LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential), which ranks Algebraic ahead of Trigonometric. Let $u = x$ (the part you differentiate) and $dv = \cos x\,dx$ (the part you integrate). 3. Compute the pieces: $du = dx$ and $v = \int \cos x\,dx = \sin x$. 4. Apply $\int u\,dv = uv - \int v\,du$: $\int x\cos x\,dx = x\sin x - \int \sin x\,dx$. 5. Finish the basic integral: $\int \sin x\,dx = -\cos x$, so $x\sin x - (-\cos x) = x\sin x + \cos x$. Add the constant of integration. Answer: $x\sin x + \cos x + C$ #### A BC-only idea: interval of convergence Problem: Find the interval of convergence of $\sum_{n=1}^{\infty} \frac{(x-2)^n}{n\cdot 3^n}$. 1. This is a power series, the heart of BC Unit 10, with nothing like it on AB. For a power series, reach for the ratio test (Topic 10.8) to find the radius, then check the endpoints by hand. 2. Write the general term $a_n = \frac{(x-2)^n}{n\cdot 3^n}$ and form the ratio $\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x-2)^{n+1}}{(n+1)\cdot 3^{n+1}} \cdot \frac{n\cdot 3^n}{(x-2)^n}\right|$. 3. Simplify. The powers of $(x-2)$ leave one factor, and $\frac{3^n}{3^{n+1}} = \frac{1}{3}$, giving $\left|\frac{x-2}{3}\right|\cdot \frac{n}{n+1}$. 4. Take the limit: $\lim_{n \to \infty} \left|\frac{x-2}{3}\right|\cdot\frac{n}{n+1} = \left|\frac{x-2}{3}\right|\cdot\lim_{n \to \infty}\frac{n}{n+1} = \left|\frac{x-2}{3}\right|\cdot 1 = \left|\frac{x-2}{3}\right|$. 5. The ratio test says the series converges when that limit is below 1: $\left|\frac{x-2}{3}\right| < 1$ means $|x-2| < 3$, so $-1 < x < 5$. The radius of convergence is 3. 6. Check each endpoint separately. At $x = 5$: $\sum \frac{3^n}{n\cdot 3^n} = \sum \frac{1}{n}$, the harmonic series, which diverges. At $x = -1$: $\sum \frac{(-3)^n}{n\cdot 3^n} = \sum \frac{(-1)^n}{n}$, the alternating harmonic series, which converges by the alternating series test. 7. Include the endpoint that converges and exclude the one that does not: the interval of convergence is $[-1, 5)$. Answer: $[-1, 5)$ ### FAQ Q: Is AP Calculus BC harder than AB? A: BC is broader, not fundamentally harder. It includes every AB topic plus two extra units, so there is more to learn and the pace is faster, but no single idea except infinite series is a big conceptual leap beyond AB's hardest material. The exams share the same format and the same difficulty per question. Q: Does BC cover everything in AB? A: Yes. Every AB topic is part of BC. BC adds integration by parts, partial fractions, improper integrals, Euler's method, logistic models, arc length, all of Unit 9 (parametric, polar, and vector-valued functions), and all of Unit 10 (sequences and series). That is why the BC exam reports a separate AB subscore. Q: Should I take AB or BC? A: Take BC if your precalculus is fast and reliable, you can handle a quicker pace, and your colleges reward the extra content with more credit. Take AB if you want to build the fundamentals at a steadier pace or found precalculus effortful. AB first, then BC over two years, is also a common route. Q: What is the AB subscore on the BC exam? A: The AB subscore is a separate 1 to 5 score the College Board reports for every BC exam, based on the questions that cover AB-level content. It lets a college see how you did on the AB core even when your overall BC score is lower, and it can earn AB-equivalent credit on its own. ## AP Calculus Exam Format: Sections, Timing, and Scoring URL: https://www.calclearn.org/guides/ap-calculus-exam-format Last updated: 2026-07-22 Answer first: The AP Calculus AB and BC exam runs 3 hours 15 minutes, built the same way for both courses. Section I is 45 multiple-choice questions in 105 minutes; Section II is 6 free-response questions in 90 minutes. Each section has a no-calculator part and a calculator part, and each section is worth half the score. ### Two sections, four parts, 3 hours 15 minutes Every AP Calculus exam, AB and BC alike, has the same shape: two sections split across four parts, timed for 3 hours 15 minutes in total. Section I is multiple choice; Section II is free response. The two courses are built the same way and differ only in the content tested, not in the format, timing, or calculator rules. | Section / part | Questions | Calculator | Time | Weight (derived) | | --- | --- | --- | --- | --- | | Section I, Part A (multiple choice) | 30 | Not permitted | 60 min | 33.3% | | Section I, Part B (multiple choice) | 15 | Required | 45 min | 16.7% | | Section II, Part A (free response) | 2 | Required | 30 min | 16.7% | | Section II, Part B (free response) | 4 | Not permitted | 60 min | 33.3% | Add the parts and the structure is symmetric. Section I holds 45 multiple-choice questions across 105 minutes; Section II holds 6 free-response questions across 90 minutes. The College Board publishes weights only at the section level: multiple choice is 50% of the score, and free response is the other 50%. The per-part percentages in the table above are not published separately; they are derived from those section weights by assuming every question within a section counts equally, so each of the 45 multiple-choice questions and each of the 6 free-response questions carries an equal share. NOTE (The two halves are equal): Section I and Section II each count for half the composite score, so a student who is strong on multiple choice but freezes on free response, or the reverse, leaves half the exam on the table. Prepare for both formats on purpose. ### What a calculator is allowed to do, and where A graphing calculator is required on exactly two of the four parts and not permitted on the other two. Section I Part B and Section II Part A allow the calculator; Section I Part A and Section II Part B do not. The no-calculator parts are the larger multiple-choice part and the final free-response part. - No calculator: Section I Part A (30 multiple-choice questions) and Section II Part B (4 free-response questions). - Graphing calculator required: Section I Part B (15 multiple-choice questions) and Section II Part A (2 free-response questions). - The two no-calculator parts together are 66.7% of the score, so most of the exam is worked by hand. NOTE (Practice the split, not just the topics): Because two thirds of the score comes from no-calculator parts, most practice should be pencil and paper so the machine never becomes a crutch. On the calculator parts, the exam expects you to write the exact integral or equation first and then evaluate it numerically; the setup earns the point even if the arithmetic slips. ### Pacing: how many minutes each question gets Divide each part's time by its question count and the exam's rhythm becomes concrete. Multiple-choice questions come fast; free-response questions are long and multi-part, so the two sections need different pacing instincts. $$\frac{60 \text{ min}}{30 \text{ questions}} = 2 \text{ min per question} \qquad \frac{45 \text{ min}}{15 \text{ questions}} = 3 \text{ min per question}$$ So Part A gives you about 2 minutes per multiple-choice question and Part B about 3 minutes, since calculator questions usually take longer to set up. Free response runs at a completely different pace. $$\frac{90 \text{ min}}{6 \text{ questions}} = 15 \text{ min per free-response question}$$ | Part | Questions | Time | Minutes per question | | --- | --- | --- | --- | | Section I, Part A | 30 | 60 min | 2.0 | | Section I, Part B | 15 | 45 min | 3.0 | | Section II, Part A | 2 | 30 min | 15.0 | | Section II, Part B | 4 | 60 min | 15.0 | NOTE (Bank time on the easy ones): Two to three minutes per multiple-choice question is not much, so move fast on the ones you recognize and flag the rest to revisit. A blank cannot earn a point, so put an answer on every multiple-choice question before time is called. ### How the 1 to 5 score is built Your raw work becomes a composite score: multiple-choice points and free-response points are combined, with each section contributing half. The College Board then maps that composite onto the reported 1 to 5 scale. That mapping is criterion referenced, not a curve. The score boundaries are set to reflect a fixed standard of what a 5, 4, 3, 2, or 1 represents, rather than to rank you against the other students who happened to sit the exam that year. The exact composite cut points shift slightly from year to year and are not published in advance, so there is no fixed number of points to memorize as the line for a 5. A 3 is the score most colleges treat as the threshold for credit, but each college sets its own policy, so check the schools you care about. Because the two sections weigh equally, points left on the free-response section cost exactly as much as missed multiple-choice questions. NOTE (What the odds look like on AB): On the May 2025 AP Calculus AB exam, 64.2% of the 286,722 students scored a 3 or higher and 20.3% earned a 5, with a mean score of 3.21. The exam is demanding but very passable with steady, weight-aware preparation. ### AB versus BC: same format, different content AB and BC sit for the same-shaped exam: the same two sections, the same four parts, the same 3 hours 15 minutes, the same calculator rules, and the same 1 to 5 scale. What changes is the content. BC covers everything on AB and adds more, so the per-unit weighting shifts and BC pulls in whole topics AB never sees. BC adds two entire units plus several topics inside the units the two courses share: - Unit 9, Parametric Equations, Polar Coordinates, and Vector-Valued Functions, worth 10-15% of the BC exam, is BC only. - Unit 10, Infinite Sequences and Series, worth 15-20% of the BC exam, is BC only. - Inside shared units, BC adds integration by parts (Topic 6.11), linear partial fractions (Topic 6.12), improper integrals (Topic 6.13), Euler's method (Topic 7.5), logistic models (Topic 7.9), and arc length (Topic 8.13). Because Units 9 and 10 claim a large share of the BC exam, the units AB and BC share are weighted more lightly on BC. Unit 1 is 10-15% of the AB multiple-choice section but 5-10% on BC, and Unit 5 falls from 15-20% on AB to 10-15% on BC. Unit 6 stays heavy on both at 15-20%. NOTE (The BC exam's AB subscore): Every BC exam also produces a separate AB subscore, a 1 to 5 score based only on the questions covering AB-level content. It reports how a BC student did on the AB core, and it can earn AB-equivalent credit on its own even when the overall BC score is lower. AB-only students do not receive a subscore. ### Worked examples #### Pacing Section I so no question goes unanswered Problem: Section I gives you 105 minutes for 45 multiple-choice questions: 30 questions in Part A with no calculator (60 minutes), then 15 questions in Part B with a calculator (45 minutes). How much time does each question get, and what pacing plan keeps you from running out before the last question? 1. Find the pace for Part A. Divide the time by the questions: 60 minutes over 30 questions is 2 minutes per question. Treat that as a ceiling, not a target; banking time on the easy questions leaves more for the hard ones. 2. Find the pace for Part B. 45 minutes over 15 questions is 3 minutes per question. Calculator questions run a little longer to set up, so the extra minute per question is expected, not slack to spend early. 3. Set checkpoints instead of watching every question. In Part A, aim to reach question 15 by the 30-minute mark and question 30 by 60 minutes. In Part B, aim to be at question 8 around the 24-minute mark. A checkpoint every stretch catches a slow start while you can still fix it. 4. Guard the last questions. When two or three minutes remain in a part, stop starting new questions and make sure every answer in that part is filled in. A blank cannot score, and a quick guess on a question you did not reach still has a chance. 5. Notice how different Section II feels. Six free-response questions in 90 minutes is 15 minutes each, a completely different rhythm: fewer, deeper questions where showing the setup and the justification is where the points are. Do not carry the multiple-choice sprint into free response. Answer: Budget about 2 minutes per no-calculator question in Part A and 3 minutes per calculator question in Part B, check your pace at questions 15 and 30 in Part A and around question 8 in Part B, and fill in every answer before time is called. Section II is a slower 15 minutes per free-response question, so switch rhythms when you reach it. ### FAQ Q: How long is the AP Calculus exam? A: 3 hours 15 minutes, the same for AB and BC. Section I (multiple choice) is 105 minutes for 45 questions, split into a 60-minute no-calculator Part A of 30 questions and a 45-minute calculator Part B of 15 questions. Section II (free response) is 90 minutes for 6 questions, split into a 30-minute calculator Part A of 2 questions and a 60-minute no-calculator Part B of 4 questions. Q: Can you use a calculator on every part of the AP Calculus exam? A: No. A graphing calculator is required on Section I Part B and Section II Part A, and not permitted on Section I Part A and Section II Part B. The two no-calculator parts together make up 66.7% of the score, so most of the exam is worked by hand. Q: Is AP Calculus BC scored differently from AB? A: No, both use the same two-section format, the same 3 hours 15 minutes, and the same 1 to 5 scale. The difference is content: BC covers all of AB plus Units 9 and 10 and several extra topics, weighted differently. Every BC exam also reports a separate AB subscore, a 1 to 5 score based on the AB-level questions, which AB-only students do not receive. Q: How is the 1 to 5 score calculated? A: Multiple-choice and free-response points combine into a single composite, with each section worth half. The College Board maps that composite onto the 1 to 5 scale using score boundaries set to a fixed standard rather than a curve, so the cut points shift slightly each year and are not published ahead of time. ## AP Calculus Calculator Policy: What Is Allowed and When URL: https://www.calclearn.org/guides/ap-calculus-calculator-policy Last updated: 2026-07-22 Answer first: A graphing calculator is required on two of the four exam parts, Section I Part B and Section II Part A, and not allowed on the other two. The exam expects four skills: plot a function, find zeros and intersections, evaluate a derivative at a point, and evaluate a definite integral numerically. ### Which parts allow a calculator The AP Calculus exam, AB and BC alike, has two sections, and each section splits into two parts: one you work with a graphing calculator and one you work without. A graphing calculator is required on Section I Part B, part of the multiple-choice section, and on Section II Part A, part of the free-response section. It is not allowed on Section I Part A, the larger multiple-choice part, or on Section II Part B, the final free-response part. NOTE (The weighting favors pencil and paper): Taken together, the two no-calculator parts are about two-thirds (66.7%) of your total score. The calculator earns points on a minority of the exam, so most of your practice should be by hand, and the machine should stay a tool you reach for on purpose rather than a crutch you cannot do without. BC follows the same calculator rules as AB. Both use the calculator on Section I Part B and Section II Part A and forbid it on the other two parts, so everything in this guide applies to either course. ### The four things your calculator must do The College Board expects fluency with four specific calculator operations, and every question on the calculator-active parts is written assuming you can perform them quickly. Anything beyond these four you should be able to handle by hand. 1. Plot the graph of a function in a viewing window you choose. 2. Find the zeros of a function (solve an equation numerically), which also locates where two graphs intersect: set one minus the other equal to zero and solve. 3. Evaluate the numerical value of a derivative at a given point. 4. Evaluate the numerical value of a definite integral. Notice what is not on the list. The calculator is not expected to find antiderivatives symbolically, factor, or display algebra steps, and on the exam you should never lean on it to. Learn your own model's exact keystrokes for these four operations before exam day, because fumbling the syntax under time pressure is where calculator-active points leak away. ### When to trust the calculator, when to do the algebra On the calculator-active parts, the numbers are usually ugly on purpose. A definite integral with no clean antiderivative, an intersection that is not a whole number, a slope that refuses to simplify: these are signals that the question wants the machine. On the no-calculator parts the numbers are built to work out by hand, so if your algebra there is turning into a mess, suspect an error rather than hunting for an arithmetic trick. Reach for the calculator when the task is compute-a-number: the total accumulated by a messy rate function, the point where two curves cross, the slope at a specific input. Do the algebra when the question says show, justify, or asks for an exact value. A calculator readout is never a justification, and a decimal is not an exact answer. NOTE (Write the setup, not just the number): On the free-response section, notation and communication are scored. A definite integral you evaluate on the calculator still has to appear on your paper as the integral, for example $\int_0^8 r(t)\,dt$, before the numeric value. A bare number with no setup earns little even when it is correct, because the reader cannot see the reasoning behind it. ### Radian mode and the mistakes that quietly cost points A handful of calculator habits separate students who gain points on the calculator parts from students who lose them, and none of them are about being clever. They are about being disciplined. NOTE (Radian mode, every time): AP Calculus works in radians. If your calculator is in degree mode, every trig value, and every derivative or integral of a function containing $\sin$, $\cos$, or $\tan$, comes out wrong. Set radian mode before the calculator section begins and confirm it, because a degree-mode slip can silently ruin an entire question. Keep full precision inside the calculator and round only at the very end. Store an intermediate result to a variable, or use the answer (Ans) key, rather than copying a rounded value back in by hand. Rounding early compounds: using $3.14$ in place of the stored value of $\pi$ throws $\pi \cdot 25^2$ off from $1963.495$ to $1962.5$, nearly a full unit of error (about $0.995$), which is more than enough to miss a three-decimal-place answer. Report final answers to three decimal places unless a question says otherwise, the standard the AP free-response instructions ask for. AP accepts an answer that is either rounded or truncated to three decimal places, so the danger is not truncation itself but reporting fewer than three places, or letting early rounding corrupt the digits before you get there. Keep three decimals and the answer stays in range. And because two of the four parts have no calculator at all, drill those skills by hand: the machine will not be there for the parts worth two-thirds of your score. ### Do not memorize an approved-model list Which calculator models are permitted changes from year to year, and the only current, authoritative list is the one the College Board publishes. Check that official calculator policy before exam day rather than trusting a list copied elsewhere, this guide included. Naming specific approved models here would only go stale. - Confirm your specific model appears on the College Board's current list of permitted calculators. - Practice on the exact calculator you plan to bring, so the four required operations are automatic instead of a menu hunt on exam day. - Carry fresh or spare batteries; a dead calculator on a calculator-active part costs you those questions outright. - Learn your model's menu path for a numeric derivative and a numeric definite integral, since the keystrokes differ by brand. ### Worked examples #### Calculator-active free response: total from a rate Problem: Section II Part A, calculator required. Water is pumped into a tank at a rate of $r(t) = 20\sin\left(\frac{t^2}{35}\right)$ gallons per minute for $0 \le t \le 8$ minutes. Find the total number of gallons pumped into the tank over the 8 minutes, and present the answer in correct free-response form. 1. Read for the operation. A total amount from a rate over an interval is an accumulation, so the tool is a definite integral of the rate: total $= \int_0^8 r(t)\,dt$. Because this is a calculator-active question and $r(t) = 20\sin\left(\frac{t^2}{35}\right)$ has no clean antiderivative, this is exactly the kind of integral the numerical integrator exists for. 2. Write the setup on paper first. Before touching the calculator, put the integral on your answer sheet: $\int_0^8 20\sin\left(\frac{t^2}{35}\right)\,dt$. The reader scores this setup, and writing it also tells you precisely what to enter. 3. Confirm radian mode, then evaluate numerically. With the calculator in radian mode, evaluate the definite integral of $20\sin\left(\frac{t^2}{35}\right)$ from $0$ to $8$ using your model's numeric-integral command. Keep every digit the calculator returns and do not round yet. 4. Round once, at the end, to three decimal places. The integrator returns $76.5703\ldots$, which rounds to $76.570$. 5. State the answer with units, and pace it. Report that the tank receives $76.570$ gallons over the 8 minutes; the integral expression together with this value is a complete response, while a lone number is not. Part A is a short two-question calculator block, so split your time evenly between the questions and resist over-polishing: once the setup is written and the value rounded, move on. Answer: The tank receives $76.570$ gallons of water over the 8 minutes, found as $\int_0^8 20\sin\left(\frac{t^2}{35}\right)\,dt \approx 76.570$. ### FAQ Q: Is a calculator required for the whole AP Calculus exam? A: No. It is required on only two of the four parts, Section I Part B and Section II Part A. Section I Part A and Section II Part B are worked without a calculator, and together those no-calculator parts are about two-thirds of your score. Q: Does BC allow a calculator on different parts than AB? A: No. AB and BC use the same calculator rules: a graphing calculator is required on Section I Part B and Section II Part A and is not allowed on the other two parts. Q: What is the single most common calculator mistake? A: Being in degree mode. AP Calculus is done in radians, so a calculator left in degree mode returns wrong values for anything involving $\sin$, $\cos$, or $\tan$, including numeric derivatives and integrals of those functions. Set and confirm radian mode before the calculator section. Q: Which calculator models are allowed? A: The list of permitted models is set by the College Board and can change from year to year, so check their current calculator policy directly rather than any secondhand list. Practice on whatever permitted model you plan to bring so the four required operations are automatic. --- # Common derivatives ## Derivative of ln x: Answer, Proof, and Mistakes URL: https://www.calclearn.org/derivative-of/ln-x Last updated: 2026-07-22 Answer first: The derivative of ln x is 1/x, valid for x > 0. In prime notation, if f(x) = ln x then f'(x) = 1/x. For a composite ln(u), the chain rule gives u'/u, not 1/u. This follows because ln x is the inverse of e^x: writing e^y = x and differentiating gives dy/dx = 1/x. $$\frac{d}{dx}\left[\ln x\right] = \frac{1}{x}$$ ### How to prove the derivative of ln x Start from $y = \ln x$. Because $\ln x$ is the inverse of the natural exponential, this is the same statement as $e^y = x$, which already forces $x > 0$ (the exponential $e^y$ is always positive). $$e^y = x$$ Differentiate both sides with respect to $x$. The right side is just 1; the left side needs the chain rule, since $y$ depends on $x$. $$e^y \cdot \frac{dy}{dx} = 1$$ Solve for $\frac{dy}{dx}$ and substitute $e^y = x$ back in. $$\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$$ NOTE (Watch the domain): $\ln x$ exists only for $x > 0$, so the derivative $\frac{1}{x}$ is stated on that domain. The related function $\ln|x|$ has derivative $\frac{1}{x}$ for every $x \neq 0$, and that is the version that appears when you antidifferentiate $\frac{1}{x}$. ### Where ln x's derivative shows up on the AP exam The rule $\frac{d}{dx}[\ln x] = \frac{1}{x}$ is introduced in Unit 2, Topic 2.7 (Derivatives of $\cos x$, $\sin x$, $e^x$, and $\ln x$). It is one of the basic derivatives the College Board expects you to recall without deriving, and Unit 2 carries a weighting of 10 to 15 percent on AB and 5 to 10 percent on BC. It most often appears inside a composite $\ln(u)$, where the chain rule (Unit 3, Topic 3.1) applies and you differentiate the inside and divide by it. It also turns up bare, or folded into a product or quotient such as $x \ln x$. $$\frac{d}{dx}\left[\ln(u)\right] = \frac{u'}{u}$$ The proof above is implicit differentiation (Topic 3.2): you write $e^y = x$, differentiate both sides, and solve for $\frac{dy}{dx}$. It works because $\ln x$ is the inverse of $e^x$, the relationship that Topic 3.3 (Differentiating Inverse Functions) is built on. That topic's own rule, $g'(x) = \frac{1}{f'(g(x))}$, reaches the same answer directly: with $f(x) = e^x$ and $g(x) = \ln x$, $g'(x) = \frac{1}{e^{\ln x}} = \frac{1}{x}$. So this one derivative ties four CED topics together. | CED topic | How the derivative of ln x appears | | --- | --- | | 2.7 Derivatives of $\cos x$, $\sin x$, $e^x$, and $\ln x$ | The rule itself, memorized in Unit 2 | | 3.1 The Chain Rule | Composites such as $\ln(x^2+1)$ | | 3.2 Implicit Differentiation | Differentiating $e^y = x$ to prove the rule | | 3.3 Differentiating Inverse Functions | The inverse-function theorem gives the same $\frac{1}{x}$ | ### Common mistakes with the derivative of ln x - Dropping the chain rule. $\frac{d}{dx}\left[\ln(x^2+1)\right] = \frac{2x}{x^2+1}$, not $\frac{1}{x^2+1}$. The inside's derivative belongs on top. - Using it where $\ln x$ is undefined. There is no value of $\ln x$ for $x \le 0$, so "$\frac{1}{x}$ for all $x$" is wrong; write $x > 0$, or $x \neq 0$ if you mean $\ln|x|$. - Confusing $\ln x$ with $e^x$. The exponential keeps itself, $\frac{d}{dx}[e^x] = e^x$, while $\frac{d}{dx}[\ln x] = \frac{1}{x}$ turns into a power. They are inverses, not the same rule. - Treating every logarithm like the natural log. For a general base, $\frac{d}{dx}\left[\log_b x\right] = \frac{1}{x \ln b}$. Only base $e$ gives $\ln b = 1$, which is why $\ln x$ produces the clean $\frac{1}{x}$. - Half-applying the chain rule on $(\ln x)^2$. The power rule and chain rule combine to $2\ln x \cdot \frac{1}{x} = \frac{2\ln x}{x}$; keep both the leftover $\ln x$ and the $\frac{1}{x}$. ### Worked chain-rule examples Example 1. Differentiate $\ln(x^2+1)$. Let $u = x^2+1$, so $u' = 2x$; apply $\frac{u'}{u}$. $$\frac{d}{dx}\left[\ln(x^2+1)\right] = \frac{2x}{x^2+1}$$ Example 2. Differentiate $\ln(\sin x)$. Now $u = \sin x$ and $u' = \cos x$, so the quotient simplifies to $\cot x$. $$\frac{d}{dx}\left[\ln(\sin x)\right] = \frac{\cos x}{\sin x} = \cot x$$ Example 3. Differentiate $\ln(5x)$. Here $u = 5x$ and $u' = 5$, and the constant cancels. $$\frac{d}{dx}\left[\ln(5x)\right] = \frac{5}{5x} = \frac{1}{x}$$ NOTE (Why the constant vanished): $\ln(5x) = \ln 5 + \ln x$, and $\ln 5$ is a constant, so $\ln(5x)$ and $\ln x$ share the same derivative $\frac{1}{x}$. Any positive constant multiplier inside the log drops out. ### FAQ Q: Is the derivative of ln x always 1/x? A: On the domain of $\ln x$, yes: for $x > 0$, $\frac{d}{dx}[\ln x] = \frac{1}{x}$. If you instead use $\ln|x|$, its derivative is $\frac{1}{x}$ for every $x \neq 0$. The rule only changes when the base is not $e$ or the inside is not just $x$. Q: What is the derivative of ln(u)? A: Use the chain rule: $\frac{d}{dx}[\ln(u)] = \frac{u'}{u}$, the derivative of the inside divided by the inside. For instance, $\frac{d}{dx}[\ln(3x^2+1)] = \frac{6x}{3x^2+1}$. Q: Why is the derivative of ln x equal to 1/x? A: Because $\ln x$ is the inverse of $e^x$. Rewrite $y = \ln x$ as $e^y = x$, differentiate both sides to get $e^y \cdot \frac{dy}{dx} = 1$, then solve: $\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$. Q: What is the derivative of log base b of x? A: $\frac{d}{dx}[\log_b x] = \frac{1}{x \ln b}$. For base 10 that is $\frac{1}{x \ln 10}$. Only the natural log, where $b = e$, gives the clean $\frac{1}{x}$, since $\ln e = 1$. ## Derivative of e^x: Answer, Proof, and Mistakes URL: https://www.calclearn.org/derivative-of/e-x Last updated: 2026-07-22 Answer first: The derivative of e^x is e^x. The natural exponential function equals its own derivative, the only function (apart from constant multiples) that does, so d/dx of e^x = e^x. This follows straight from the limit that defines the number e. Do not use the power rule: e^x is not x times e^(x-1). $$\frac{d}{dx}\left[e^x\right] = e^x$$ ### Why e^x equals its own derivative Start from the limit definition of the derivative and factor. Because $e^{x+h}=e^{x}\cdot e^{h}$, the fixed factor $e^{x}$ pulls straight out of the limit, and everything that depends on $h$ collects into one small piece. $$\frac{d}{dx}e^{x}=\lim_{h \to 0}\frac{e^{x+h}-e^{x}}{h}=\lim_{h \to 0}\frac{e^{x}\left(e^{h}-1\right)}{h}=e^{x}\lim_{h \to 0}\frac{e^{h}-1}{h}$$ So the whole derivative rides on one limit: $\lim_{h \to 0}\frac{e^{h}-1}{h}$. The number $e$ is defined to be the exact base that makes this limit equal to $1$ (equivalently, the base whose graph has slope $1$ at $x=0$). Substituting that value finishes the proof. $$\lim_{h \to 0}\frac{e^{h}-1}{h}=1 \quad\Longrightarrow\quad \frac{d}{dx}e^{x}=e^{x}\cdot 1=e^{x}$$ Because $\frac{d}{dx}e^{x}=e^{x}$, differentiating never changes it, and neither does antidifferentiating. It is the only function, apart from constant multiples $Ce^{x}$, with that property: any $y$ satisfying $y'=y$ must equal $Ce^{x}$. This is what separates $e^{x}$ from its look-alikes. In $x^{e}$ the variable is the base and the exponent is the constant $e \approx 2.718$, so the power rule applies and gives $e\,x^{e-1}$. In a general exponential $a^{x}$ the derivative is $a^{x}\ln a$; $e^{x}$ is simply the case $a=e$, where $\ln e=1$ and the extra factor disappears. | Function | Type | Rule that applies | Derivative | | --- | --- | --- | --- | | $e^{x}$ | exponential, base $e$ | equals its own derivative | $e^{x}$ | | $a^{x}$ | exponential, base $a>0$ | multiply by $\ln a$ | $a^{x}\ln a$ | | $x^{e}$ | power, constant exponent | power rule | $e\,x^{e-1}$ | ### Where e^x shows up on the AP exam The rule first appears in CED Topic 2.7, Derivatives of cos x, sin x, e^x, and ln x, which names $e^{x}$ outright. That topic sits in Unit 2 (Differentiation: Definition and Fundamental Properties), and Unit 2 carries 10 to 15 percent of the AB exam weighting and 5 to 10 percent on BC. Most exam appearances are composites, not the bare $e^{x}$. Once a function like $e^{3x}$ or $e^{-x^{2}}$ shows up, you layer the chain rule from CED Topic 3.1 (Unit 3, Differentiation: Composite, Implicit, and Inverse Functions) on top of the base rule. $$\frac{d}{dx}\,e^{g(x)}=e^{g(x)}\cdot g'(x)$$ - Composite derivatives such as $e^{kx}$, $e^{-x^{2}}$, and $e^{\sin x}$, handled with the chain rule (Topic 3.1). - Product and quotient combinations like $x^{2}e^{x}$ or $\frac{e^{x}}{x}$, where $e^{x}$ is one factor (Topics 2.8 and 2.9). - Exponential growth and decay solutions $y=Ce^{kt}$ from separable differential equations (Unit 7). - Tangent-line, linearization, and L'Hopital's Rule problems where you evaluate $e^{x}$ and its derivative at a point (Unit 4). NOTE: The self-replicating behavior runs backward too: $\int e^{x}\,dx=e^{x}+C$ (CED Topic 6.8). The natural exponential is its own antiderivative as well as its own derivative, which is why $e^{x}$ survives every round of calculus you do to it. ### Common mistakes Almost every lost point on this rule traces back to one of four specific errors. Each has a clean fix once you name what went wrong. - Using the power rule. The exponent is the variable, so $e^{x}$ does not become $x\,e^{x-1}$. The power rule is only for a constant exponent on a variable base, like $x^{e}$. - Dropping the inside derivative. $\frac{d}{dx}e^{3x}=3e^{3x}$, not $e^{3x}$; the chain rule factor of $3$ is part of the answer. - Losing the sign on $e^{-x}$. The inside derivative is $-1$, so $\frac{d}{dx}e^{-x}=-e^{-x}$. - Over-generalizing to other bases. $\frac{d}{dx}2^{x}=2^{x}\ln 2$, not $2^{x}$. Only base $e$ drops the $\ln a$ factor, because $\ln e=1$. NOTE: Quick self-check: if your answer to $\frac{d}{dx}e^{x}$ has an $x$ in it, you used the power rule by mistake. The correct answer is exactly $e^{x}$, with no coefficient and no shifted exponent. ### Chain rule composites, worked The recipe for any $e^{g(x)}$ is the same: copy $e^{g(x)}$ down unchanged, then multiply by the derivative of the exponent. Two standard exam composites make the pattern concrete. $$\frac{d}{dx}\,e^{3x}=e^{3x}\cdot\frac{d}{dx}(3x)=e^{3x}\cdot 3=3e^{3x}$$ The second example has a polynomial inside, so the multiplier is no longer a constant. It appears constantly in normal-distribution and other statistics setups. $$\frac{d}{dx}\,e^{-x^{2}}=e^{-x^{2}}\cdot\frac{d}{dx}\left(-x^{2}\right)=e^{-x^{2}}\cdot(-2x)=-2x\,e^{-x^{2}}$$ NOTE: One more pattern worth memorizing: for $e^{\sin x}$ the outside stays $e^{\sin x}$ and you multiply by $\cos x$, giving $\frac{d}{dx}e^{\sin x}=\cos x \cdot e^{\sin x}$. Same recipe, trig inside. ### FAQ Q: Is the derivative of e^x really just e^x? A: Yes. $e^{x}$ is the one function, up to constant multiples, that equals its own derivative. It falls out of the limit definition: $\frac{d}{dx}e^{x}=e^{x}\lim_{h\to 0}\frac{e^{h}-1}{h}$, and $e$ is defined precisely so that limit is $1$. Q: What is the derivative of e^(2x) or e^(kx)? A: Use the chain rule: $\frac{d}{dx}e^{kx}=k\,e^{kx}$, so $\frac{d}{dx}e^{2x}=2e^{2x}$. The outside $e^{kx}$ stays put, and you multiply by the derivative of the exponent. Q: Why can't I use the power rule on e^x? A: The power rule $\frac{d}{dx}x^{n}=nx^{n-1}$ works only when the variable is the base and the exponent is constant. In $e^{x}$ the base is the constant $e$ and the variable is in the exponent, so it is exponential, not a power. Writing $x\,e^{x-1}$ is wrong. Q: How is the derivative of e^x different from a^x? A: For any base, $\frac{d}{dx}a^{x}=a^{x}\ln a$. When $a=e$, $\ln e=1$, so the factor drops and $\frac{d}{dx}e^{x}=e^{x}$. That missing $\ln a$ is exactly what makes $e$ the natural base. ## Derivative of sin x: Answer, Proof, Mistakes URL: https://www.calclearn.org/derivative-of/sin-x Last updated: 2026-07-22 Answer first: The derivative of sin x is cos x, for every real number x, provided x is measured in radians. It comes from the limit definition using the two special limits sin(h)/h -> 1 and (cos h - 1)/h -> 0. For a composite like sin(3x), the chain rule gives 3 cos(3x). $$\frac{d}{dx}\left[\sin x\right] = \cos x$$ ### Proof from the limit definition Start from the limit definition of the derivative and expand $\sin(x+h)$ with the angle-addition identity. Two special limits do all the work, so the proof is really about isolating them. $$f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$$ The identity $\sin(x+h) = \sin x \cos h + \cos x \sin h$ lets you substitute and then group the terms that contain $\sin x$ apart from the terms that contain $\cos x$. $$\lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} = \lim_{h \to 0} \left[ \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h} \right]$$ Because $x$ is held fixed, $\sin x$ and $\cos x$ are constants with respect to $h$ and pull outside the limit. Now apply the two special limits. The first, $\lim_{h \to 0} \frac{\sin h}{h} = 1$, comes from the Squeeze Theorem in Topic 1.8; the second, $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$, follows from it algebraically. $$\lim_{h \to 0} \frac{\sin h}{h} = 1, \qquad \lim_{h \to 0} \frac{\cos h - 1}{h} = 0$$ Substituting these two values collapses the whole expression to a single term. $$f'(x) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x$$ NOTE: Radians only. Both special limits hold only when $h$ is in radians. In degrees, $\lim_{h \to 0} \frac{\sin h}{h} = \frac{\pi}{180}$, so the derivative would carry that factor: $\frac{d}{dx}[\sin x] = \frac{\pi}{180}\cos x$. Every AP Calculus formula assumes radians, so keep your calculator in radian mode. ### Where sin x derivatives show up on the AP exam The derivative of $\sin x$ is introduced in Topic 2.7, Derivatives of cos x, sin x, e^x, and ln x, inside Unit 2, Differentiation: Definition and Fundamental Properties. Unit 2 carries 10-15% of the AB exam and 5-10% of the BC exam, and $\sin x$ rarely appears bare. It almost always sits inside a larger structure that decides which rule you reach for. - Chain rule composites (Topic 3.1): $\sin(3x)$, $\sin(x^2)$, or $\sin(e^x)$, where you multiply $\cos(\text{inner})$ by the inner derivative. - Product and quotient rule (Topics 2.8 and 2.9): expressions like $x^2 \sin x$ or $\frac{\sin x}{x}$ need $\frac{d}{dx}[\sin x] = \cos x$ as one of the pieces. - Tables of values and graphical reasoning: free-response parts that ask for $f'$ at a point when $f$ involves $\sin x$. - Motion and related rates (Unit 4): position functions with a sine term give velocity through this derivative. - BC series (Topic 10.14): the Maclaurin series of $\sin x$ depends on the repeating derivative pattern $\cos x$, $-\sin x$, $-\cos x$, $\sin x$. ### Common mistakes to avoid 1. Adding a wrong sign. The derivative of $\sin x$ is $+\cos x$, with no minus sign. The minus sign belongs to $\frac{d}{dx}[\cos x] = -\sin x$. Mixing up which one is negative is the single most common error. 2. Dropping the chain rule on a composite. $\frac{d}{dx}[\sin(3x)] = 3\cos(3x)$, not $\cos(3x)$. The inner derivative is a factor you cannot skip. 3. Confusing $\sin(x^2)$ with $(\sin x)^2$. The first is a sine of a composite, derivative $2x\cos(x^2)$; the second is a square of sine, derivative $2\sin x \cos x$. They are different functions with different rules. 4. Working in degrees. If your calculator is in degree mode, a numerical derivative will be off by a factor of $\frac{\pi}{180}$. The formula $\cos x$ is a radians statement. 5. Treating $\sin x$ like a power. There is no power rule here; you do not bring down an exponent. Use the memorized derivative $\cos x$. ### Chain rule composites: two worked examples Both examples follow the same recognition step: the outer function is $\sin(\text{something})$, so the derivative starts as $\cos(\text{something})$, then multiplies by the derivative of the inside. Example 1. Differentiate $\sin(3x^2)$. The inside is $3x^2$, whose derivative is $6x$. $$\frac{d}{dx}\left[\sin(3x^2)\right] = \cos(3x^2) \cdot 6x = 6x\cos(3x^2)$$ Example 2. Differentiate $\sin(e^x)$. The inside is $e^x$, whose derivative is again $e^x$. $$\frac{d}{dx}\left[\sin(e^x)\right] = \cos(e^x) \cdot e^x = e^x\cos(e^x)$$ In both cases the $\cos$ keeps the original inside untouched, and the extra factor is exactly the inner derivative. Never simplify the inside of the cosine while you are still taking the derivative. ### FAQ Q: Why is the derivative of sin x equal to cos x and not -cos x? A: The limit-definition proof produces $\sin x \cdot 0 + \cos x \cdot 1$, which is $+\cos x$ with no minus sign. The minus sign shows up instead for $\frac{d}{dx}[\cos x] = -\sin x$, because there the surviving term is $-\sin x \cdot 1$. Q: Does the formula only work in radians? A: Yes. The proof relies on $\lim_{h \to 0} \frac{\sin h}{h} = 1$, which is only true when $h$ is in radians. In degrees that limit equals $\frac{\pi}{180}$, so the derivative would become $\frac{\pi}{180}\cos x$. AP Calculus always uses radians. Q: What is the derivative of sin(2x)? A: Apply the chain rule. The outer derivative is $\cos(2x)$ and the inner derivative of $2x$ is $2$, so $\frac{d}{dx}[\sin(2x)] = 2\cos(2x)$. Q: What is the second derivative of sin x? A: Differentiate twice: $\frac{d}{dx}[\sin x] = \cos x$, then $\frac{d}{dx}[\cos x] = -\sin x$. So the second derivative is $-\sin x$. The derivatives cycle every four steps through $\cos x$, $-\sin x$, $-\cos x$, $\sin x$. ## Derivative of cos x: Answer, Proof, Mistakes URL: https://www.calclearn.org/derivative-of/cos-x Last updated: 2026-07-22 Answer first: The derivative of cos x is -sin x. Cosine's derivative always carries a minus sign. It comes straight from the limit definition using the angle-addition formula, and it holds only when x is measured in radians. When cosine wraps an inner function, multiply by that inner derivative using the chain rule. $$\frac{d}{dx}\left[\cos x\right] = -\sin x$$ ### Where the minus sign comes from The derivative of cosine is one of four you memorize in Topic 2.7, but you never have to take it on faith. It drops out of the limit definition the moment you expand $\cos(x+h)$ with the angle-addition formula. $$f'(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}$$ Replace $\cos(x+h)$ with $\cos x\cos h - \sin x\sin h$, then group the two terms that share a factor of $\cos x$. $$= \lim_{h \to 0} \left(\cos x \cdot \frac{\cos h - 1}{h} - \sin x \cdot \frac{\sin h}{h}\right)$$ Two standard limits finish it. As $h \to 0$, $\frac{\sin h}{h} \to 1$ is the squeeze-theorem limit from Topic 1.8, and $\frac{\cos h - 1}{h} \to 0$ follows from it. The first term collapses to zero and the second leaves the answer. $$f'(x) = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$$ The minus sign is not decoration. At $x = 0$ cosine sits at its peak, so its slope is $0$; just to the right the curve falls, so the slope must be negative there. $-\sin x$ does exactly that, since $-\sin 0 = 0$ and $-\sin x < 0$ for small $x > 0$. NOTE: Shift intuition: differentiating a sine or cosine slides its graph left by $\frac{\pi}{2}$. Because $\cos x = \sin\left(x + \frac{\pi}{2}\right)$, one more shift lands on $-\sin x$. The pattern cycles every four derivatives: $\cos x \to -\sin x \to -\cos x \to \sin x \to \cos x$. ### Where it shows up on the AP exam Cosine's derivative is introduced in Topic 2.7 (Derivatives of cos x, sin x, e^x, and ln x), part of Unit 2, which carries 10-15% of the AB exam and 5-10% of BC. Cosine rarely appears bare, though. It nearly always shows up wrapped around an inner function, so it travels with the chain rule (Topic 3.1). Two settings account for most of its exam appearances: - Motion: with position $x(t) = \cos(\omega t)$, velocity is $-\omega\sin(\omega t)$, and that sign flip is what sends the object back the other way (Topic 4.2). - Tables of values: given $g(x) = \cos(u(x))$ and a table of $u$ and $u'$, you evaluate $g'(a) = -\sin(u(a)) \cdot u'(a)$ at a specific point. These reward knowing the minus sign cold. This kind of recall is expected on sight, so a dropped minus sign is a self-inflicted point loss on a problem you otherwise know how to do. ### Common mistakes Nearly every error traces back to the minus sign or to a forgotten inner function. | Common slip | Correct | | --- | --- | | $\frac{d}{dx}[\cos x] = \sin x$ | $\frac{d}{dx}[\cos x] = -\sin x$ | | $\frac{d}{dx}[\cos(3x)] = 3\sin(3x)$ | $\frac{d}{dx}[\cos(3x)] = -3\sin(3x)$ | | $\frac{d}{dx}[\cos(x^2)] = -\sin(x^2)$ | $\frac{d}{dx}[\cos(x^2)] = -2x\sin(x^2)$ | | $\int \cos x\,dx = -\sin x$ | $\int \cos x\,dx = \sin x + C$ | The first row is the classic sign flip. The second drops the minus after applying the chain rule. The third forgets to multiply by the inner derivative $2x$. The fourth borrows the derivative's minus sign for the integral, where it does not belong: differentiating cosine adds a minus, while integrating cosine does not. NOTE: Radians only. The rule $-\sin x$ assumes $x$ is in radians. In degrees the derivative would be $-\frac{\pi}{180}\sin x$, because $\frac{\sin h}{h} \to 1$ holds only for radian measure. AP Calculus is in radians throughout, so this bites mainly when a calculator is left in degree mode. ### Two chain-rule composites Once the table entry is automatic, the only new work is the inner derivative. Write $-\sin(\text{inner})$ first, then multiply by the derivative of the inner function. Example 1. Differentiate $\cos(3x)$. The inner function is $3x$, whose derivative is $3$. $$\frac{d}{dx}[\cos(3x)] = -\sin(3x) \cdot 3 = -3\sin(3x)$$ Example 2. Differentiate $\cos^2 x$, which means $(\cos x)^2$. This stacks the power rule on top of the cosine rule: bring down the $2$, keep $\cos x$, then multiply by the derivative of $\cos x$. $$\frac{d}{dx}[\cos^2 x] = 2\cos x \cdot (-\sin x) = -2\sin x\cos x = -\sin(2x)$$ The last step uses the double-angle identity $2\sin x\cos x = \sin(2x)$; either $-2\sin x\cos x$ or $-\sin(2x)$ is a correct final answer. ### FAQ Q: Is the derivative of cos x sin x or negative sin x? A: It is $-\sin x$. Cosine differentiates to negative sine, and the minus sign is the part students drop most often. Sine is the one without the minus: $\frac{d}{dx}[\sin x] = \cos x$. Q: Why does the derivative of cos x have a minus sign? A: Because cosine is falling wherever sine is positive. Near $x = 0$ cosine leaves its maximum and decreases, so its slope must be negative there, and $-\sin x$ supplies exactly that sign. The limit-definition proof makes it precise through the identity $\cos(x+h) = \cos x\cos h - \sin x\sin h$. Q: What is the derivative of cos x squared? A: For $\cos^2 x = (\cos x)^2$, the power and chain rules give $\frac{d}{dx}[\cos^2 x] = 2\cos x \cdot (-\sin x) = -\sin(2x)$. Do not confuse this with $\cos(x^2)$, whose derivative is $-2x\sin(x^2)$. Q: What is the second derivative of cos x? A: Differentiate twice: $\frac{d}{dx}[\cos x] = -\sin x$, then $\frac{d}{dx}[-\sin x] = -\cos x$. So the second derivative of $\cos x$ is $-\cos x$, which is why $y = \cos x$ satisfies $y'' = -y$. ## Derivative of tan x: Answer, Proof, and Mistakes URL: https://www.calclearn.org/derivative-of/tan-x Last updated: 2026-07-22 Answer first: The derivative of tan x is sec^2 x, which equals 1/cos^2 x. It holds everywhere tan x is defined, meaning every x except odd multiples of pi/2 (where cos x = 0). You get it by applying the quotient rule to tan x = sin x / cos x, using the Pythagorean identity sin^2 x + cos^2 x = 1. $$\frac{d}{dx}\left[\tan x\right] = \sec^2 x$$ ### The proof, by the quotient rule Rewrite the tangent as a quotient of sine and cosine, then differentiate with the quotient rule. Recall that $\frac{d}{dx}[\sin x] = \cos x$ and $\frac{d}{dx}[\cos x] = -\sin x$. $$\frac{d}{dx}\left[\tan x\right] = \frac{d}{dx}\left[\frac{\sin x}{\cos x}\right] = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}$$ The two negatives in $-\sin x \cdot (-\sin x)$ make a positive, so the numerator becomes $\cos^2 x + \sin^2 x$, which equals $1$ by the Pythagorean identity. $$= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$$ NOTE (Same answer, two forms): $\sec^2 x$ means $(\sec x)^2$, and since $\sec x = \frac{1}{\cos x}$, that is exactly $\frac{1}{\cos^2 x}$. Both forms name the same result. It holds only where $\cos x \neq 0$, so it excludes $x = \frac{\pi}{2} + n\pi$, the odd multiples of $\frac{\pi}{2}$ where $\tan x$ is undefined. ### Where it shows up on the AP exam The College Board lists this derivative directly in Unit 2, Topic 2.10, Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions. Unit 2, Differentiation: Definition and Fundamental Properties, makes up 10 to 15% of the AB exam and 5 to 10% of BC. The derivation leans on Topic 2.9, The Quotient Rule, and the ingredients come from Topic 2.7, Derivatives of cos x, sin x, e^x, and ln x. On the exam you rarely differentiate a bare $\tan x$; it usually sits inside a composite that needs the chain rule from Unit 3, Topic 3.1. - Chain-rule composites such as $\tan(3x)$ or $\tan(x^2)$, where the answer is $\sec^2(u)$ times $u'$ for inside function $u$. - Related rates and motion problems where an angle changes, so a $\tan$ term is differentiated with respect to time. - Tangent-line and slope questions that ask for $f'$ at a point when $f$ contains $\tan x$. - Free-response work that expects you to recognize $\sec^2 x$ instantly, without pausing to re-derive it. ### Common mistakes | Mistake | Why it is wrong | | --- | --- | | Writing $\frac{d}{dx}[\tan x] = \sec x \tan x$ | That is the derivative of $\sec x$, not $\tan x$. The tangent gives $\sec^2 x$. | | $\frac{d}{dx}[\tan(3x)] = \sec^2(3x)$ | The chain rule is missing. The inner derivative makes it $3\sec^2(3x)$. | | Reading $\sec^2 x$ as $\sec(x^2)$ | $\sec^2 x$ means $(\sec x)^2$, a squared output, not a squared input. | | Numerator $\cos^2 x - \sin^2 x$ | A sign slip. Since $\frac{d}{dx}[\cos x] = -\sin x$, the term is $-\sin x \cdot (-\sin x) = +\sin^2 x$, so the sum gives $1$. | | Using the formula at $x = \frac{\pi}{2}$ | Both $\tan x$ and $\sec^2 x$ are undefined there. Check the domain before you evaluate. | ### Chain-rule composites, quick practice Every composite follows one pattern: differentiate the outer tangent to $\sec^2(u)$, then multiply by $u'$, the derivative of the inside function $u$. $$\frac{d}{dx}\left[\tan(5x)\right] = \sec^2(5x) \cdot 5 = 5\sec^2(5x)$$ $$\frac{d}{dx}\left[\tan(x^2 + 1)\right] = \sec^2(x^2 + 1) \cdot 2x = 2x\sec^2(x^2 + 1)$$ For a product such as $x\tan x$, pair this with the product rule: $\frac{d}{dx}[x\tan x] = \tan x + x\sec^2 x$. ### FAQ Q: Is the derivative of $\tan x$ $\sec^2 x$ or $\frac{1}{\cos^2 x}$? A: Both, because they are the same expression. Since $\sec x = \frac{1}{\cos x}$, squaring gives $\sec^2 x = \frac{1}{\cos^2 x}$. Use whichever form fits the rest of your work. Q: Why is the derivative of $\tan x$ always positive? A: Wherever it is defined, $\sec^2 x = \frac{1}{\cos^2 x}$ is $1$ divided by a nonzero square, so it is always positive, never zero or negative. And since $\cos^2 x \leq 1$, dividing $1$ by it gives $\sec^2 x \geq 1$. Either way the slope is positive, which is the calculus reason $\tan x$ is increasing on every interval between its asymptotes. Q: What is the second derivative of $\tan x$? A: Differentiate $\sec^2 x$ with the chain rule: $\frac{d}{dx}[(\sec x)^2] = 2\sec x \cdot \sec x \tan x = 2\sec^2 x \tan x$. Q: Where is $\tan x$ not differentiable? A: At the odd multiples of $\frac{\pi}{2}$, that is $x = \frac{\pi}{2} + n\pi$ for any integer $n$. There $\cos x = 0$, so $\tan x$ has a vertical asymptote and is not even defined, let alone differentiable. ## Derivative of sec x: Answer, Proof, and Mistakes URL: https://www.calclearn.org/derivative-of/sec-x Last updated: 2026-07-22 Answer first: The derivative of sec x is sec x tan x. Write sec x as (cos x)^-1 and apply the chain rule: the outer power gives -(cos x)^-2 and the inner derivative of cos x is -sin x, so the two negatives cancel and leave sin x over cos^2 x, which regroups as sec x tan x. $$\frac{d}{dx}\left[\sec x\right] = \sec x \tan x$$ ### Proving it with the chain rule Start from the definition of secant. Since $\sec x = \frac{1}{\cos x} = (\cos x)^{-1}$, the derivative is just a power of a function, so the chain rule (Topic 3.1) applies directly. The quotient rule (Topic 2.9) on $\frac{1}{\cos x}$ lands on the same result if you prefer that route. $$\frac{d}{dx}\left[(\cos x)^{-1}\right] = -1\cdot(\cos x)^{-2}\cdot(-\sin x) = \frac{\sin x}{\cos^2 x}$$ The outer power rule contributes one negative sign and the derivative of the inner $\cos x$ contributes another, so the two cancel. Now regroup the single fraction into a product: $$\frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x \tan x$$ That cancellation is exactly why the derivative of $\sec x$ carries no leading minus sign, while its co-function $\csc x$ does. ### The co-function sign pattern Notice there is no minus sign in $\sec x \tan x$. Set it beside its co-function: $\csc x$ differentiates to $-\csc x \cot x$. Across all six trig derivatives, the three co-functions (cosine, cotangent, cosecant) are exactly the ones whose derivatives carry a leading negative sign. | Function $f(x)$ | Derivative $f'(x)$ | Leading sign | | --- | --- | --- | | $\sec x$ | $\sec x \tan x$ | $+$ | | $\csc x$ | $-\csc x \cot x$ | $-$ | So the sec and csc rules are mirror images. Keep $\sec x \tan x$ in memory, then swap each factor for its co-function and attach a minus sign to recover $\csc x \to -\csc x \cot x$. The full six-row table lives in the trig derivatives guide. ### Where sec x shows up on the AP exam The secant derivative is introduced in Unit 2 (Differentiation: Definition and Fundamental Properties), specifically Topic 2.10, Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions. Unit 2 carries 10-15% of the AB exam and 5-10% of BC. On the exam $\sec x$ rarely stands alone. It usually appears inside a composite such as $\sec(2x)$ or $\sec(x^2)$, which pairs Topic 2.10 with the chain rule (Topic 3.1, Unit 3). It also turns up in tangent-line approximation (Unit 4, Topic 4.6) and related-rates work (Unit 4), where $\sec$ and $\tan$ show up together. ### Common mistakes to avoid - Answering $\sec^2 x$. That is the derivative of $\tan x$, not of $\sec x$. Keep the pair straight: $\tan x \to \sec^2 x$ and $\sec x \to \sec x \tan x$. - Attaching a negative sign. The derivative of $\sec x$ has no leading minus sign; the minus belongs to its co-function, $\csc x \to -\csc x \cot x$. - Dropping the inner derivative on composites. $\frac{d}{dx}\sec(3x) = 3\sec(3x)\tan(3x)$, not $\sec(3x)\tan(3x)$. The chain rule factor is required. - Writing only one factor. The answer is the product $\sec x \cdot \tan x$; leaving off either $\sec x$ or $\tan x$ loses half of it. - Treating $\sec x$ as $\frac{1}{\sec x}$ or as an inverse function. Recall $\sec x = \frac{1}{\cos x}$, so its derivative comes from differentiating $(\cos x)^{-1}$, not from a reciprocal or an arcsecant rule. ### Two worked chain-rule composites Every composite runs off one template: $\frac{d}{dx}\sec u = \sec u \tan u \cdot u'$, where $u'$ is the derivative of whatever sits inside. Linear inside. Let $u = 3x$, so $u' = 3$. $$\frac{d}{dx}\sec(3x) = \sec(3x)\tan(3x)\cdot 3 = 3\sec(3x)\tan(3x)$$ Power inside. Let $u = x^2$, so $u' = 2x$. $$\frac{d}{dx}\sec(x^2) = \sec(x^2)\tan(x^2)\cdot 2x = 2x\sec(x^2)\tan(x^2)$$ ### FAQ Q: Is the derivative of sec x equal to sec^2 x? A: No. $\sec^2 x$ is the derivative of $\tan x$. The derivative of $\sec x$ is $\sec x \tan x$, the product of secant and tangent. The two are easy to swap, so lock in the pair $\tan x \to \sec^2 x$ and $\sec x \to \sec x \tan x$. Q: Why does the derivative of sec x have no minus sign, but csc x does? A: In the proof, the power rule and the derivative of $\cos x$ each contribute a minus sign, and the two cancel, so no minus sign survives. For $\csc x = (\sin x)^{-1}$ the inner derivative is $+\cos x$, so nothing cancels and a leading minus sign remains: $-\csc x \cot x$. This is the co-function sign pattern. Q: How do you differentiate sec(2x) or other composites? A: Use the chain rule. Differentiate the outside as $\sec(\text{inside})\tan(\text{inside})$, then multiply by the derivative of the inside. For $\sec(2x)$ the inside derivative is $2$, giving $2\sec(2x)\tan(2x)$. Q: Where is the derivative of sec x undefined? A: At $x = \frac{\pi}{2} + n\pi$ (the odd multiples of $\frac{\pi}{2}$), because $\cos x = 0$ there. At those points $\sec x$ itself is undefined, so neither the function nor its derivative $\sec x \tan x$ exists. ## Derivative of sqrt(x): Answer, Proof, and Mistakes URL: https://www.calclearn.org/derivative-of/sqrt-x Last updated: 2026-07-22 Answer first: The derivative of the square root of x is 1 over 2 times the square root of x, written f'(x) = 1/(2 sqrt(x)). It comes from the power rule: rewrite sqrt(x) as x^(1/2), so the derivative is (1/2)x^(-1/2). The slope is defined for x > 0 and grows without bound as x approaches 0. $$\frac{d}{dx}\left[\sqrt{x}\right] = \frac{1}{2\sqrt{x}}$$ ### The proof: it is just the power rule The square root is a power in disguise. Rewrite $\sqrt{x}$ as $x^{1/2}$, then apply the power rule $\frac{d}{dx}\left[x^n\right] = n\,x^{n-1}$ with $n = \frac{1}{2}$. $$\frac{d}{dx}\left[x^{1/2}\right] = \frac{1}{2}\,x^{1/2 - 1} = \frac{1}{2}\,x^{-1/2} = \frac{1}{2}\cdot\frac{1}{\sqrt{x}} = \frac{1}{2\sqrt{x}}$$ NOTE (Rewrite before you differentiate): Radicals and variables in a denominator are power-rule problems in disguise. Turn every square root into a fractional exponent, $\sqrt{x} = x^{1/2}$, and every $\frac{1}{x^k}$ into $x^{-k}$, before you touch the derivative. Skipping that step is where most coefficient and sign errors begin. ### Why the slope blows up at x = 0 The function $\sqrt{x}$ is defined at $x = 0$, where $\sqrt{0} = 0$, but its derivative is not. As $x$ shrinks toward $0$, $\sqrt{x}$ shrinks with it, so $\frac{1}{2\sqrt{x}}$ grows without bound. $$\lim_{x \to 0^+} \frac{1}{2\sqrt{x}} = +\infty$$ Geometrically, the graph leaves the origin with a vertical tangent line, and a vertical line has no finite slope. So $\sqrt{x}$ is continuous at $0$ but not differentiable there. The domain of $f'(x) = \frac{1}{2\sqrt{x}}$ is $x > 0$, one endpoint narrower than the domain $x \ge 0$ of $\sqrt{x}$ itself. ### Where √x shows up on the AP exam You rarely differentiate a bare $\sqrt{x}$ on the exam. The plain case is a one-line power-rule check, but square roots earn their points inside larger expressions, where they force a chain rule or hide in a distance formula. | Context | What you do | CED location | | --- | --- | --- | | A bare $\sqrt{x}$ term | Power rule with $n = \frac{1}{2}$ | Unit 2, Topic 2.5 (Applying the Power Rule) | | A composite $\sqrt{g(x)}$ | Chain rule: outside times inside | Unit 3, Topic 3.1 (The Chain Rule) | | Distance $\sqrt{x^2 + y^2}$ in related rates | Chain rule while differentiating with respect to time | Unit 4, Topics 4.4 to 4.5 | | Arc length integrand, $\sqrt{1 + [f'(x)]^2}$ (BC) | Set up the integral, then evaluate | Unit 8, Topic 8.13 | The differentiation units are heavily weighted. Unit 2 (fundamental derivative rules) is worth 10 to 15 percent of the AB exam and 5 to 10 percent of BC; Unit 3 (chain rule and composites) is worth 5 to 10 percent on both. Knowing $\frac{d}{dx}\left[\sqrt{x}\right]$ cold frees time for the harder composites built on top of it. ### Common mistakes - Misplacing the coefficient $\frac{1}{2}$. The derivative is $\frac{1}{2\sqrt{x}}$, not $2\sqrt{x}$ or $\frac{\sqrt{x}}{2}$. The coefficient $\frac{1}{2}$ multiplies $x^{-1/2}$, so both the $2$ and the root land in the denominator, never in the numerator. - Forgetting to subtract $1$ from the exponent. Differentiating $x^{1/2}$ gives $\frac{1}{2}x^{-1/2}$, not $\frac{1}{2}x^{1/2}$. The new exponent is $\frac{1}{2} - 1 = -\frac{1}{2}$. - Dropping the chain rule. For $\sqrt{g(x)}$ the answer is $\frac{g'(x)}{2\sqrt{g(x)}}$; writing $\frac{1}{2\sqrt{g(x)}}$ alone leaves out the inside derivative $g'(x)$. - Reporting a slope at $x = 0$. There is none, since the tangent is vertical, so $f'(0)$ does not exist even though $\sqrt{0} = 0$. - Confusing the derivative with the integral. $\frac{d}{dx}\left[\sqrt{x}\right] = \frac{1}{2\sqrt{x}}$, while $\int \sqrt{x}\,dx = \frac{2}{3}x^{3/2} + C$. Adding $1$ to the exponent is integration; subtracting $1$ is differentiation. ### Two chain-rule composites Every composite $\sqrt{g(x)}$ reuses the same outside derivative, $\frac{1}{2\sqrt{g(x)}}$, multiplied by the inside derivative $g'(x)$. Take $\frac{d}{dx}\left[\sqrt{x^2 + 1}\right]$. Here $g(x) = x^2 + 1$ and $g'(x) = 2x$. $$\frac{d}{dx}\left[\sqrt{x^2 + 1}\right] = \frac{1}{2\sqrt{x^2 + 1}}\cdot 2x = \frac{x}{\sqrt{x^2 + 1}}$$ Take $\frac{d}{dx}\left[\sqrt{\sin x}\right]$. Here $g(x) = \sin x$ and $g'(x) = \cos x$. $$\frac{d}{dx}\left[\sqrt{\sin x}\right] = \frac{1}{2\sqrt{\sin x}}\cdot \cos x = \frac{\cos x}{2\sqrt{\sin x}}$$ The pattern never changes: differentiate the root, then multiply by the derivative of whatever sits under it. ### FAQ Q: What is the derivative of $\sqrt{x}$? A: It is $\frac{1}{2\sqrt{x}}$, valid for $x > 0$. Rewrite $\sqrt{x}$ as $x^{1/2}$ and use the power rule: the exponent $\frac{1}{2}$ comes down as a coefficient and drops to $-\frac{1}{2}$, giving $\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$. Q: Why is the $2$ in the denominator? A: Because the power rule brings the exponent $\frac{1}{2}$ down as a coefficient. That $\frac{1}{2}$ multiplies $x^{-1/2} = \frac{1}{\sqrt{x}}$, so the product is $\frac{1}{2}\cdot\frac{1}{\sqrt{x}} = \frac{1}{2\sqrt{x}}$, with both the $2$ and the root below the bar. Q: Is $\sqrt{x}$ differentiable at $x = 0$? A: No. The function is continuous at $0$ because $\sqrt{x} \to 0$ as $x \to 0^+$, matching its value $\sqrt{0} = 0$; but the graph has a vertical tangent there, so the slope is undefined. The derivative $\frac{1}{2\sqrt{x}}$ exists only for $x > 0$. Q: How do you differentiate $\sqrt{g(x)}$? A: Use the chain rule: $\frac{d}{dx}\left[\sqrt{g(x)}\right] = \frac{g'(x)}{2\sqrt{g(x)}}$. For example, $\frac{d}{dx}\left[\sqrt{3x}\right] = \frac{3}{2\sqrt{3x}}$ and $\frac{d}{dx}\left[\sqrt{x^2 + 1}\right] = \frac{x}{\sqrt{x^2 + 1}}$. ## Derivative of 1/x: Answer, Proof, and Mistakes URL: https://www.calclearn.org/derivative-of/1-over-x Last updated: 2026-07-22 Answer first: The derivative of 1/x is -1/x^2, also written -x^(-2). Rewrite 1/x as x^(-1) and use the power rule: the exponent -1 moves out front and drops by one, giving -1 times x^(-2) = -1/x^2. Because x^2 is positive for every x except 0, this slope is always negative, so 1/x decreases on each of its two branches. $$\frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^2}$$ ### The proof: rewrite 1/x as a power Start by writing the reciprocal as a power. Since $\frac{1}{x} = x^{-1}$, this is an ordinary power-rule derivative, not a separate rule to memorize. $$\frac{d}{dx}\left[x^{-1}\right] = (-1)\,x^{-1-1} = -x^{-2} = -\frac{1}{x^2}$$ The power rule says $\frac{d}{dx}\left[x^n\right] = n\,x^{n-1}$. With $n = -1$, the exponent $-1$ comes down as the coefficient out front, and the new exponent is $-1 - 1 = -2$. That leading $-1$ is the source of the minus sign in the answer. The quotient rule gives the same result. Treating $\frac{1}{x}$ as a quotient with constant numerator $1$: $$\frac{d}{dx}\left[\frac{1}{x}\right] = \frac{x\cdot 0 - 1\cdot 1}{x^2} = -\frac{1}{x^2}$$ The derivative of the numerator is $0$, so only the $-1\cdot 1$ term survives, and the $x^2$ in the denominator comes from squaring the bottom. ### The slope is negative everywhere, so 1/x is decreasing For every $x \neq 0$, $x^2 > 0$, so $-\frac{1}{x^2}$ is negative. The slope of $\frac{1}{x}$ is never zero and never positive, which means it has no critical points and no turning points. So $\frac{1}{x}$ is decreasing on each of its two branches, $(-\infty, 0)$ and $(0, \infty)$, taken separately. Avoid saying it is decreasing for all $x$: the domain is split by the vertical asymptote at $x = 0$, where the function is undefined, and the two branches are not one connected piece. | $x$ | $\frac{1}{x}$ | $\frac{d}{dx}\left[\frac{1}{x}\right]$ | | --- | --- | --- | | $-2$ | $-\frac{1}{2}$ | $-\frac{1}{4}$ | | $-\frac{1}{2}$ | $-2$ | $-4$ | | $\frac{1}{2}$ | $2$ | $-4$ | | $2$ | $\frac{1}{2}$ | $-\frac{1}{4}$ | Every entry in the last column is negative. The slope grows steeper (more negative) as $x$ approaches $0$ and flattens toward $0$ as $|x|$ gets large. ### Where 1/x shows up on the AP exam On the AP Calculus AB and BC exams, $x^{-1}$ is a direct application of the power rule from Unit 2, Topic 2.5 (Applying the Power Rule). Unit 2 carries about 10-15% of the AB exam and 5-10% of the BC exam. It shows up more often inside composites. Whenever you differentiate $\frac{1}{g(x)}$, the chain rule from Unit 3, Topic 3.1 (The Chain Rule) turns the outer $\frac{1}{u}$ into $-\frac{1}{u^2}$ and multiplies by $g'(x)$. Differentiating $\frac{1}{g(x)}$ is also a special case of the quotient rule (Topic 2.9), with a constant numerator, so you get the same $-\frac{1}{g(x)^2}$ factor either way. NOTE (Do not confuse it with the antiderivative): $\frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^2}$, but $\int \frac{1}{x}\,dx = \ln|x| + C$ (Unit 6, Topic 6.8). Differentiating $\frac{1}{x}$ and antidifferentiating it point in opposite directions, and neither one equals $\ln x$. ### Common mistakes - Dropping the minus sign and writing $\frac{1}{x^2}$ instead of $-\frac{1}{x^2}$. The derivative is negative everywhere, so a positive answer is wrong on sight. - Mixing up the two directions: $\frac{d}{dx}\left[\ln x\right] = \frac{1}{x}$, while $\frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^2}$. The reciprocal $\frac{1}{x}$ being the derivative of $\ln x$ does not make it the derivative of $\frac{1}{x}$. - Botching the new exponent: from $x^{-1}$ you subtract $1$ to reach $x^{-2}$, not $x^{0}$ or $x^{-1}$. Subtracting from a negative exponent is where the slip happens. - Assuming the derivative comes from squaring the function. The $x^2$ in the denominator comes from the power rule (or from squaring the denominator in the quotient rule), not from squaring $\frac{1}{x}$. - Skipping the chain rule on composites: $\frac{d}{dx}\left[\frac{1}{3x+1}\right]$ is not $-\frac{1}{(3x+1)^2}$, because you still multiply by the inner derivative $3$. ### Two chain-rule examples Example 1. Differentiate $\frac{1}{3x+1}$. Let $u = 3x + 1$, so the outer function is $\frac{1}{u}$ with derivative $-\frac{1}{u^2}$, and the inner derivative is $u' = 3$. $$\frac{d}{dx}\left[\frac{1}{3x+1}\right] = -\frac{1}{(3x+1)^2}\cdot 3 = -\frac{3}{(3x+1)^2}$$ Example 2. Differentiate $\frac{1}{x^2+1}$. Here $u = x^2 + 1$ and $u' = 2x$, so multiply the outer $-\frac{1}{u^2}$ by $2x$. $$\frac{d}{dx}\left[\frac{1}{x^2+1}\right] = -\frac{1}{(x^2+1)^2}\cdot 2x = -\frac{2x}{(x^2+1)^2}$$ In both problems the outer step is the same $-\frac{1}{u^2}$ you get from $\frac{1}{x}$; the chain rule just appends the inner derivative. ### FAQ Q: What is the derivative of 1/x? A: It is $-\frac{1}{x^2}$. Rewrite $\frac{1}{x}$ as $x^{-1}$ and apply the power rule: the exponent $-1$ multiplies out front and drops to $-2$, giving $-x^{-2} = -\frac{1}{x^2}$. Q: Why is the derivative of 1/x negative? A: Because $-\frac{1}{x^2}$ has $x^2 > 0$ in the denominator for every $x \neq 0$, the whole expression stays negative. That matches the graph: $\frac{1}{x}$ falls as you move left to right along each branch. Q: Is the derivative of 1/x the same as the derivative of ln x? A: No. $\frac{d}{dx}\left[\ln x\right] = \frac{1}{x}$, but $\frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^2}$. The reciprocal $\frac{1}{x}$ is the derivative of $\ln x$; differentiating $\frac{1}{x}$ itself gives a different, negative result. Q: What is the second derivative of 1/x? A: Differentiate $-\frac{1}{x^2} = -x^{-2}$ again: $\frac{d}{dx}\left[-x^{-2}\right] = 2x^{-3} = \frac{2}{x^3}$. So the second derivative of $\frac{1}{x}$ is $\frac{2}{x^3}$. ## Derivative of a^x: Answer, Proof, and Mistakes URL: https://www.calclearn.org/derivative-of/a-x Last updated: 2026-07-22 Answer first: The derivative of a^x is a^x times ln a, for any positive constant base a. So d/dx of 2^x is 2^x ln 2. It comes from rewriting a^x as e^(x ln a) and differentiating with the chain rule; when a = e the factor is ln e = 1, so the extra factor disappears, which is why e^x is its own derivative. $$\frac{d}{dx}\left[a^x\right] = a^x \ln a$$ ### The proof: rewrite the base as a power of e The $a^x$ rule is not a separate fact to memorize. Every positive number is a power of $e$, since $a = e^{\ln a}$. Raise both sides to the $x$ and the exponents multiply, which turns any exponential into a base-$e$ exponential with a linear inner function. $$a^x = \left(e^{\ln a}\right)^x = e^{x \ln a}$$ Now differentiate $e^{x \ln a}$ with the chain rule. The outside is $e^u$, whose derivative is $e^u$; the inside is $u = x \ln a$, and because $\ln a$ is a constant, $\frac{du}{dx} = \ln a$. $$\frac{d}{dx}\, e^{x \ln a} = e^{x \ln a} \cdot \ln a = a^x \ln a$$ The base comes back down unchanged and a constant factor of $\ln a$ drops out front. That factor is the whole story: it is why every exponential except one carries an extra number when you differentiate it. The lone exception is base $e$. Set $a = e$ and the factor becomes $\ln e = 1$, so $a^x \ln a$ collapses to $e^x \cdot 1 = e^x$. That is the real reason $e^x$ is its own derivative, and the reason $e$ is the natural base to build on. NOTE (The one line to remember): $\frac{d}{dx}\, a^x = a^x \ln a$, valid for any constant base $a > 0$. The base rides along unchanged and $\ln a$ appears as a constant factor. Only base $e$ escapes the extra factor, because $\ln e = 1$. ### a^x is not x^a: the base-versus-exponent trap The most common wrong turn is reaching for the power rule. The power rule $\frac{d}{dx}\, x^n = n x^{n-1}$ applies when the variable is the base and the exponent is a constant. In $a^x$ the roles are swapped: the base is constant and the variable is the exponent, so the power rule does not apply at all. Read the position of $x$ before you pick a rule. | Function | Where the variable sits | Rule | Derivative | | --- | --- | --- | --- | | $x^a$ | in the base | power rule | $a x^{a-1}$ | | $a^x$ | in the exponent | exponential rule | $a^x \ln a$ | | $x^x$ | in both | logarithmic differentiation | $x^x(\ln x + 1)$ | The $x^x$ row is a reminder that when the variable appears in both places, neither rule alone works and you take $\ln$ of both sides first. That case is past what $a^x$ needs, but it shows why naming the base and the exponent out loud, before writing anything, is worth the two seconds it costs. ### Where a^x shows up on the AP exam The CED lists the derivatives you are expected to know cold in Topic 2.7, "Derivatives of cos x, sin x, e^x, and ln x." Notice $a^x$ is not on that list. The exam expects you to build it from two things that are: the $e^x$ rule in Topic 2.7 and the chain rule in Topic 3.1. Rewriting $a^x$ as $e^{x \ln a}$ is the bridge between them. Unit 2 is 10-15% of the AB exam and 5-10% of BC; Unit 3, which owns the chain rule, is 5-10% of both. Topic 3.5, "Selecting Procedures for Calculating Derivatives," is where the exam checks that you recognize an exponential base and reach for $a^x \ln a$ rather than the power rule. In practice $a^x$ arrives most often inside a composite such as $a^{g(x)}$, or in a table-of-values problem that hands you values of a function $f$ and asks for the derivative of $2^{f(x)}$ at a point. Both reduce to the same rule plus one chain-rule factor. ### Common mistakes - Using the power rule: writing $\frac{d}{dx}\, a^x = x\, a^{x-1}$. This misapplies the power rule. The power rule is for $x^a$, where the variable sits in the base, and gives $a x^{a-1}$; here the variable is in the exponent, so the exponential rule applies instead. - Dropping the $\ln a$ factor: writing $\frac{d}{dx}\, a^x = a^x$. That is true only for $a = e$. Every other base leaves a constant factor of $\ln a$ out front. - Forgetting the inner derivative on a composite: $\frac{d}{dx}\, 3^{x^2}$ is not $3^{x^2} \ln 3$. The chain rule multiplies by $\frac{d}{dx}\, x^2 = 2x$. - Confusing $\ln a$ with $\ln x$: the factor is $\ln$ of the constant base, a fixed number, not a function of $x$. For $2^x$ it is the constant $\ln 2 \approx 0.693$. ### Two worked composites with the chain rule Both use the same composite form. Writing $F(x) = a^{g(x)}$ gives $F'(x) = a^{g(x)} \ln a \cdot g'(x)$: the base rule times the derivative of the exponent. Example 1. Differentiate $3^{x^2}$. Here $F(x) = 3^{x^2}$, the base is $3$, the inner function is $g(x) = x^2$, and $g'(x) = 2x$. $$F'(x) = 3^{x^2} \ln 3 \cdot 2x = 2x \ln 3 \cdot 3^{x^2}$$ Example 2. Differentiate $5^{\sin x}$. Here $F(x) = 5^{\sin x}$, the base is $5$, the inner function is $g(x) = \sin x$, and $g'(x) = \cos x$. $$F'(x) = 5^{\sin x} \ln 5 \cdot \cos x$$ In each case the base stays put, $\ln$ of the base rides in front as a constant, and the chain rule tacks on the derivative of the exponent. If you can reproduce those three moves in order, the general exponential is no harder than $e^x$. ### FAQ Q: What is the derivative of 2^x? A: $\frac{d}{dx}\, 2^x = 2^x \ln 2$, which is the general rule $\frac{d}{dx}\, a^x = a^x \ln a$ with $a = 2$. Numerically $\ln 2 \approx 0.693$. Writing $x\, 2^{x-1}$ is the power rule and does not apply, because the variable is in the exponent. Q: Why does a^x keep a factor of ln a but e^x does not? A: Both follow the same rule, $a^x \ln a$. For base $e$ that factor is $\ln e$, which equals $1$, so it disappears and $e^x$ differentiates to itself. Base $e$ is simply the base whose natural log is $1$. Q: How is the derivative of a^x different from x^a? A: $x^a$ has the variable in the base, so it uses the power rule: $\frac{d}{dx}\, x^a = a x^{a-1}$. $a^x$ has the variable in the exponent, so it uses the exponential rule: $\frac{d}{dx}\, a^x = a^x \ln a$. Same symbols, opposite rules. Q: How do I differentiate a^x with a chain rule, like 4^(3x)? A: Multiply by the derivative of the exponent: $\frac{d}{dx}\, 4^{3x} = 4^{3x} \ln 4 \cdot 3 = 3 \ln 4 \cdot 4^{3x}$. In general, for $F(x) = a^{g(x)}$, $F'(x) = a^{g(x)} \ln a \cdot g'(x)$. ## Derivative of x^x: Answer, Proof, and Mistakes URL: https://www.calclearn.org/derivative-of/x-x Last updated: 2026-07-22 Answer first: The derivative of x^x is x^x(ln x + 1), valid for x > 0. Because x^x has the variable in both the base and the exponent, neither the power rule nor the exponential rule works alone. Taking the natural log of both sides and differentiating implicitly (logarithmic differentiation) gives the result. $$\frac{d}{dx}\left[x^x\right] = x^x(\ln x + 1)$$ ### Why neither the power rule nor the exponential rule works The function $x^x$ puts the variable in two places at once: it is both the base and the exponent. Every basic derivative rule assumes one of those two is a fixed number, so no single rule applies here. The power rule $\frac{d}{dx}\left[x^n\right] = n\,x^{n-1}$ needs the exponent $n$ to be constant, but in $x^x$ the exponent is the variable $x$. The exponential rule $\frac{d}{dx}\left[a^x\right] = a^x \ln a$ needs the base $a$ to be constant, but in $x^x$ the base is also the variable $x$. Each rule breaks on the part the other one handles. NOTE (The two wrong answers add up to the right one): Force the power rule and you get $x \cdot x^{x-1} = x^x$. Force the exponential rule and you get $x^x \ln x$. The correct derivative $x^x(\ln x + 1)$ is exactly the sum of those two wrong answers. That is a useful way to remember the result, and the proof below shows why it happens. ### Proof by logarithmic differentiation Since no single rule fits, take the natural log of both sides first. That turns the stubborn exponent into a product, which the product rule can handle. This technique is called logarithmic differentiation. Start by naming the function $y = x^x$ and applying $\ln$ to both sides. $$y = x^x \quad\Longrightarrow\quad \ln y = \ln\left(x^x\right) = x \ln x$$ Differentiate both sides with respect to $x$. The left side needs the chain rule because $y$ is a function of $x$, and the right side $x \ln x$ needs the product rule. $$\frac{1}{y}\cdot\frac{dy}{dx} = (1)\cdot\ln x + x\cdot\frac{1}{x} = \ln x + 1$$ Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$, then replace $y$ with $x^x$. $$\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$$ NOTE (Domain: x must be positive): As a continuous real function, $x^x = e^{x \ln x}$, and that $\ln x$ forces $x > 0$. The same $\ln x$ shows up in the proof step $\ln y = x \ln x$, so both the function and this method require positive $x$. The derivative $x^x(\ln x + 1)$ holds on that same domain, and there is no real derivative at $x \le 0$. ### Where this shows up on the AP exam You rarely see $x^x$ named directly on the AP exam, but every move in its derivative is a tested skill. The $\ln x$ derivative in the middle step is Topic 2.7 (Derivatives of $\cos x$, $\sin x$, $e^x$, and $\ln x$). Differentiating $\ln y$ as a function of $x$ is implicit differentiation, Topic 3.2, and the factor $\frac{dy}{dx}$ it produces comes from the chain rule, Topic 3.1. The larger skill is noticing that a standard rule does not fit and choosing a different route, which is exactly Topic 3.5, Selecting Procedures for Calculating Derivatives. A function with a variable in both the base and the exponent is the signal to reach for logarithmic differentiation instead of the power or exponential rule. ### Common mistakes - Using the power rule and writing $x \cdot x^{x-1} = x^x$. That treats the exponent as constant and drops the entire $\ln x$ term. - Using the exponential rule and writing $x^x \ln x$. That treats the base as constant and drops the $+1$. - Forgetting the chain rule on the left side: writing $\frac{d}{dx}\left[\ln y\right] = \frac{1}{y}$ instead of $\frac{1}{y}\cdot\frac{dy}{dx}$. Without that $\frac{dy}{dx}$ factor you can never solve for the derivative. - Reporting the answer for every $x$. As a continuous real function, $x^x$ and its derivative exist only for $x > 0$. - In a composite such as $e^{x^x}$ or $\left(x^x\right)^2$, forgetting to multiply by the inner derivative $x^x(\ln x + 1)$. ### Quick chain-rule composites Once you have $\frac{d}{dx}\left[x^x\right] = x^x(\ln x + 1)$, it becomes the inner piece of any composite. Here are two. Composite 1: differentiate $e^{x^x}$. The outer function is $e^u$ with $u = x^x$, so the chain rule gives $e^u$ times the derivative of the inside. $$\frac{d}{dx}\left[e^{x^x}\right] = e^{x^x}\cdot\frac{d}{dx}\left[x^x\right] = e^{x^x}\cdot x^x(\ln x + 1)$$ Composite 2: differentiate $\ln\left(x^x\right)$ as a self-check. By the chain rule it is $\frac{1}{x^x}$ times the derivative of $x^x$. $$\frac{d}{dx}\left[\ln\left(x^x\right)\right] = \frac{1}{x^x}\cdot x^x(\ln x + 1) = \ln x + 1$$ Because $\ln\left(x^x\right) = x \ln x$, differentiating directly with the product rule also gives $\ln x + 1$. The two routes agree, which confirms the derivative of $x^x$. As a numeric anchor, at $x = 1$ the slope is $1^1(\ln 1 + 1) = 1$, so $y = x^x$ passes through $(1, 1)$ with slope $1$. ### FAQ Q: What is the derivative of x^x? A: It is $x^x(\ln x + 1)$, valid for $x > 0$. You find it by logarithmic differentiation: set $y = x^x$, take $\ln$ of both sides to get $\ln y = x \ln x$, differentiate implicitly to get $\frac{1}{y}\cdot\frac{dy}{dx} = \ln x + 1$, then multiply by $y = x^x$. Q: Why can't I use the power rule on x^x? A: The power rule $\frac{d}{dx}\left[x^n\right] = n\,x^{n-1}$ only works when the exponent $n$ is a fixed number. In $x^x$ the exponent is the variable $x$, so the rule does not apply. Forcing it gives $x^x$, which is missing the $\ln x$ term of the real answer. Q: What is the domain of the derivative of x^x? A: As a continuous real function, $x^x = e^{x \ln x}$, whose $\ln x$ needs a positive input, so both $x^x$ and its derivative are defined only for $x > 0$. The proof step $\ln y = x \ln x$ inherits the same restriction, and there is no real derivative at $x \le 0$. Q: Is x^x the same as e^(x ln x)? A: Yes. Rewriting $x^x = e^{x \ln x}$ gives a second valid method: differentiate the exponent $x \ln x$ to get $\ln x + 1$, then multiply by $e^{x \ln x} = x^x$, which returns the same $x^x(\ln x + 1)$. ## Derivative of arctan x: Answer, Proof, Mistakes URL: https://www.calclearn.org/derivative-of/arctan-x Last updated: 2026-07-22 Answer first: The derivative of arctan x is 1/(1+x^2). It holds for every real number x, since arctangent is differentiable everywhere. For a composite arctan(u), the chain rule gives u'/(1+u^2). You get it by differentiating tan y = x implicitly and using the identity 1 + tan^2 y = sec^2 y. $$\frac{d}{dx}\left[\arctan x\right] = \frac{1}{1+x^2}$$ ### The proof by implicit differentiation Start from the definition of arctangent. If $y = \arctan x$, then $\tan y = x$, with $y$ restricted to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ so that the inverse is single-valued. Differentiate both sides with respect to $x$. The left side needs the chain rule, since $y$ is itself a function of $x$: $$\sec^2 y \cdot \frac{dy}{dx} = 1$$ Solve for $\frac{dy}{dx}$, then rewrite $\sec^2 y$ with the Pythagorean identity $1 + \tan^2 y = \sec^2 y$: $$\frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y}$$ Because $\tan y = x$, replace $\tan^2 y$ with $x^2$. Every trig piece cancels out and the result is a plain rational function: $$\frac{dy}{dx} = \frac{1}{1 + x^2}$$ NOTE (Why no trig is left): The identity does the work. $\sec^2 y$ still hides an angle $y$ you cannot evaluate, but $1 + \tan^2 y$ trades it for $1 + x^2$. That swap is why every inverse trig derivative comes out algebraic, with no arctangent or secant remaining. ### The chain rule form, with two worked composites On the AP exam the input is almost never a bare $x$. Whenever arctangent wraps a function $u$, multiply by that inner derivative: $$\left(\arctan u\right)' = \dfrac{u'}{1 + u^2}$$ First composite: $u = 3x$, so $u' = 3$. $$\left(\arctan(3x)\right)' = \dfrac{3}{1 + (3x)^2} = \dfrac{3}{1 + 9x^2}$$ Second composite: $u = x^2$, so $u' = 2x$. $$\left(\arctan\left(x^2\right)\right)' = \dfrac{2x}{1 + \left(x^2\right)^2} = \dfrac{2x}{1 + x^4}$$ Square the whole input before adding 1: $(3x)^2 = 9x^2$ and $\left(x^2\right)^2 = x^4$. Copying $u$ straight into the denominator without squaring it is the fastest way to lose the point. ### Where it shows up on the AP exam Inverse trig derivatives have their own place in the Course and Exam Description: Unit 3, Topic 3.4, Differentiating Inverse Trigonometric Functions. Unit 3 is worth 5-10% of both the AB and the BC exam. Arctangent's derivative is a clean rational function with no square root, which makes it a convenient inverse for both differentiation and antidifferentiation problems. You will usually meet it inside a chain-rule composite (Topic 3.1) or folded into a product or quotient on the no-calculator section. The reverse direction matters just as much: since $\left(\arctan x\right)' = \dfrac{1}{1+x^2}$, the antiderivative $\int \frac{1}{1+x^2}\,dx = \arctan x + C$ is a basic antiderivative rule in Unit 6, and completing the square (Topic 6.10) turns many integrals into exactly that arctangent form. NOTE (One fact, two directions): Learn $\left(\arctan x\right)' = \dfrac{1}{1+x^2}$ and you also own $\int \frac{1}{1+x^2}\,dx = \arctan x + C$. Free-response scoring rewards students who spot a $1 + x^2$ in a denominator and reach for arctangent. ### Common mistakes - Squaring the wrong thing. The denominator is $1 + x^2$, a sum, not $(1 + x)^2$. There is no cross term and nothing to expand. - Adding a stray negative sign. Arctangent's derivative is positive, $\frac{1}{1+x^2}$. The minus sign belongs to its cousins: $\left(\operatorname{arccot} x\right)' = -\dfrac{1}{1+x^2}$ and $\left(\arccos x\right)' = -\dfrac{1}{\sqrt{1-x^2}}$. - Confusing it with arcsine. $\left(\arcsin x\right)' = \dfrac{1}{\sqrt{1-x^2}}$ has a square root over a difference; arctangent has neither. Arctangent's derivative is defined for every real $x$, while arcsine's needs $|x| < 1$. - Dropping the chain rule. $\left(\arctan(5x)\right)' = \dfrac{5}{1+25x^2}$, not $\dfrac{1}{1+25x^2}$. The inner derivative multiplies the entire fraction. ### FAQ Q: Is the derivative of $\arctan x$ the same as the derivative of $\tan^{-1} x$? A: Yes. $\arctan x$ and $\tan^{-1} x$ are two notations for the same inverse tangent function, so both have derivative $\frac{1}{1+x^2}$. Watch the notation trap: here $\tan^{-1} x$ means the inverse function, not $(\tan x)^{-1} = \cot x$, which is unrelated. Q: Why does the trig disappear from the answer? A: Implicit differentiation first gives $\frac{1}{\sec^2 y}$, which still hides the angle $y$. The identity $\sec^2 y = 1 + \tan^2 y$ lets you substitute $\tan y = x$, so $\sec^2 y = 1 + x^2$. What is left is a rational function with no arctangent or secant in it. Q: What is the domain of the derivative of $\arctan x$? A: All real numbers. Since $1 + x^2 \geq 1$ for every $x$, the denominator is never zero, so $\frac{1}{1+x^2}$ is defined everywhere. That is different from $\arcsin x$, whose derivative $\frac{1}{\sqrt{1-x^2}}$ only exists for $|x| < 1$. Q: What is the second derivative of $\arctan x$? A: Differentiate $\frac{1}{1+x^2} = \left(1+x^2\right)^{-1}$ with the chain rule: $\left(\left(1+x^2\right)^{-1}\right)' = -\left(1+x^2\right)^{-2} \cdot 2x = \dfrac{-2x}{\left(1+x^2\right)^2}$. It is zero at $x = 0$, the inflection point of $\arctan x$. ## Derivative of arcsin x: Answer, Proof, Mistakes URL: https://www.calclearn.org/derivative-of/arcsin-x Last updated: 2026-07-22 Answer first: The derivative of arcsin x is 1/sqrt(1 - x^2), valid for -1 < x < 1. In prime notation, if f(x) = arcsin x then f'(x) = 1/sqrt(1 - x^2). The proof differentiates sin(f(x)) = x implicitly and simplifies cos back to sqrt(1 - x^2) on arcsin's range. The derivative of arccos x is the same but negative. $$\frac{d}{dx}\left[\arcsin x\right] = \frac{1}{\sqrt{1-x^2}}$$ ### How to prove the derivative of arcsin x Start from $y = \arcsin x$. By the definition of the inverse sine, this is the same statement as $\sin y = x$ with $y$ restricted to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the range that keeps $\arcsin$ a genuine function. $$\sin y = x$$ Differentiate both sides with respect to $x$. The right side is just 1; the left side needs the chain rule, since $y$ depends on $x$. $$\cos y \cdot \frac{dy}{dx} = 1$$ Solve for $\frac{dy}{dx}$, then rewrite $\cos y$ in terms of $x$. From $\sin y = x$ and the identity $\sin^2 y + \cos^2 y = 1$, you get $\cos y = \pm\sqrt{1 - x^2}$. $$\frac{dy}{dx} = \frac{1}{\cos y}, \qquad \cos y = \pm\sqrt{1 - x^2}$$ The range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ settles the sign: cosine is never negative on that interval, so take the positive root. Substituting gives the derivative. $$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$$ NOTE (Watch the domain and the endpoints): $\arcsin x$ is defined on the closed interval $[-1, 1]$, but the derivative needs $1 - x^2 > 0$, so $\frac{1}{\sqrt{1-x^2}}$ exists only on the open interval $-1 < x < 1$. At $x = \pm 1$ the denominator is 0: the graph of $\arcsin$ has vertical tangents there, and the derivative is undefined. NOTE (arccos is the mirror image): The same proof on $\cos y = x$ gives $\frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}}$, the exact negative. It has to be: $\arcsin x + \arccos x = \frac{\pi}{2}$ is constant, so differentiating both sides forces their derivatives to cancel. ### Where arcsin x's derivative shows up on the AP exam The rule $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$ is introduced in Unit 3, Topic 3.4 (Differentiating Inverse Trigonometric Functions). Unit 3, Differentiation: Composite, Implicit, and Inverse Functions, is weighted 5 to 10 percent on both the AB and BC exams. The proof above is Topic 3.2 (Implicit Differentiation) applied to $\sin y = x$, and it is a special case of Topic 3.3 (Differentiating Inverse Functions). On the exam the rule is rarely asked bare; it usually appears wrapped in the chain rule (Topic 3.1) on a composite $\arcsin(u)$. $$\frac{d}{dx}\left[\arcsin(u)\right] = \frac{u'}{\sqrt{1 - u^2}}$$ It also runs in reverse. Recognizing that $\frac{1}{\sqrt{1-x^2}}$ antidifferentiates back to $\arcsin x$ is part of choosing an antidifferentiation technique in Unit 6 (Topic 6.14), so this one derivative pays off in both the differentiation and integration units. | CED topic | How the derivative of arcsin x appears | | --- | --- | | 3.4 Differentiating Inverse Trigonometric Functions | The rule itself, $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$ | | 3.2 Implicit Differentiation | The $\sin y = x$ derivation | | 3.1 The Chain Rule | Composites such as $\arcsin(2x)$ and $\arcsin(x^2)$ | | 6.14 Selecting Techniques for Antidifferentiation | Reverse: $\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C$ | ### Common mistakes with the derivative of arcsin x - Putting the minus sign on arcsin. The derivative of $\arcsin x$ is $+\frac{1}{\sqrt{1-x^2}}$; the negative belongs to $\frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}}$. Same denominator, opposite sign. - Reading $\arcsin x$ as $\frac{1}{\sin x}$. The notation $\sin^{-1} x$ means the inverse sine, not a reciprocal. The reciprocal $\frac{1}{\sin x} = \csc x$ is a different function with derivative $-\csc x \cot x$. - Forgetting to square the inside on a composite. For $\arcsin(u)$ the denominator is $\sqrt{1 - u^2}$, so $\frac{d}{dx}[\arcsin(x^3)] = \frac{3x^2}{\sqrt{1 - x^6}}$, with $x^6 = (x^3)^2$ under the root, not $x^3$. - Using the derivative at $x = \pm 1$. Those endpoints are in the domain of $\arcsin$, but the derivative blows up there (vertical tangents), so $\frac{1}{\sqrt{1-x^2}}$ is valid only on $-1 < x < 1$. - Dropping the chain rule factor. $\frac{d}{dx}[\arcsin(2x)] = \frac{2}{\sqrt{1-4x^2}}$, not $\frac{1}{\sqrt{1-4x^2}}$; the inner derivative 2 multiplies, and the inside becomes $(2x)^2 = 4x^2$. ### Worked chain-rule examples Every composite $\arcsin(u)$ runs on the same template: $\frac{1}{\sqrt{1-u^2}}$ times $u'$. Track the inside $u$ carefully, because it plays two roles at once: it is squared under the root and differentiated on top. Example 1. Differentiate $\arcsin(2x)$. Here $u = 2x$, so $u' = 2$ and $u^2 = 4x^2$. $$\frac{d}{dx}\left[\arcsin(2x)\right] = \frac{2}{\sqrt{1 - 4x^2}}$$ Example 2. Differentiate $\arcsin(x^2)$. Now $u = x^2$, so $u' = 2x$ and $u^2 = x^4$. $$\frac{d}{dx}\left[\arcsin(x^2)\right] = \frac{2x}{\sqrt{1 - x^4}}$$ NOTE (Keep the two roles of u separate): In Example 2 the inside $x^2$ shows up twice and differently: squared under the root it becomes $x^4$, differentiated on top it becomes $2x$. Mixing those two roles (writing $x^2$ under the root, or $2x^2$ on top) is the usual slip. ### FAQ Q: Why is the derivative of arcsin x equal to 1/sqrt(1-x^2)? A: Because $\arcsin x$ is the inverse of sine on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Rewrite $y = \arcsin x$ as $\sin y = x$, differentiate to get $\cos y \cdot \frac{dy}{dx} = 1$, then use $\cos y = \sqrt{1-x^2}$ (positive on that range) to get $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$. Q: What is the derivative of arccos x? A: It is $-\frac{1}{\sqrt{1-x^2}}$, the negative of the arcsin derivative. Since $\arcsin x + \arccos x = \frac{\pi}{2}$ is a constant, differentiating both sides shows the two derivatives must be exact opposites. Q: What is the derivative of arcsin(u)? A: Use the chain rule: $\frac{d}{dx}[\arcsin(u)] = \frac{u'}{\sqrt{1-u^2}}$, the derivative of the inside over the square root of one minus the inside squared. For example, $\frac{d}{dx}[\arcsin(3x)] = \frac{3}{\sqrt{1-9x^2}}$. Q: Where is the derivative of arcsin x undefined? A: At $x = \pm 1$. The denominator $\sqrt{1-x^2}$ is 0 there, which matches the vertical tangents on the graph of $\arcsin$, so the derivative exists only on the open interval $-1 < x < 1$ even though $\arcsin x$ itself is defined on $[-1, 1]$. --- # Reference tables ## Derivative Rules: The Complete Table URL: https://www.calclearn.org/reference/derivative-rules Last updated: 2026-07-22 Answer first: The core rules are constant (derivative 0), power (x^n to n x^(n-1)), constant multiple, sum, difference, product, quotient, and chain. On top are the six trig derivatives, the inverse trig derivatives, and the exponential and log rules. Pick a rule by the function's outermost operation. ### How to pick the right rule Choosing a derivative rule is a recognition task, not a guessing game. Look at the outermost operation: the last thing you would do if you plugged in a number. That single question routes you to the right rule almost every time. The table below maps what a function looks like to the rule it needs, and the sections after it give the formulas grouped so the patterns are easy to remember. NOTE (The one question that decides): Identify the outermost operation. A sum wants the sum rule, a product wants the product rule, a composition wants the chain rule. Only after that do you differentiate the inside pieces. | When the function looks like | Use this rule | Why / watch for | | --- | --- | --- | | A number alone, like $7$ or $\pi$ | Constant | A constant never changes, so its slope is $0$. | | A single power, like $x^5$ or $x^{-2}$ | Power | Works for any real exponent, including negatives and roots. | | Terms joined by $+$ or $-$ | Sum and difference | Differentiate each term on its own, then recombine. | | A number times a function, like $4\cos x$ | Constant multiple | The coefficient rides along: the derivative is $-4\sin x$. | | Two functions multiplied | Product | Not $f'g'$: each factor pairs with the other's derivative. | | One function over another | Quotient | First check whether a rewrite turns it into a sum of powers. | | A function nested inside another, like $\sin(x^2)$ | Chain | Outside derivative times inside derivative; keep the inside. | These are Topics 2.5, 2.6, 2.8, 2.9, and 3.1 in the CED. The trig, inverse trig, exponential, and log rows below plug into these structural rules: whenever an inside function is more than a bare $x$, the chain rule multiplies by that inside function's derivative. ### Structural rules (Topics 2.5, 2.6, 2.8, 2.9, 3.1) These seven rules combine or transform other functions. They are the scaffolding: the specific function derivatives in the later tables get fed into them. | Rule | Formula | When to use it | | --- | --- | --- | | Constant | $\dfrac{d}{dx}[c] = 0$ | A number standing alone. Its rate of change is zero. | | Power | $\dfrac{d}{dx}[x^{n}] = n x^{n-1}$ | Any single power of $x$, including negative and fractional exponents. The workhorse for polynomials and radicals. | | Constant multiple | $\dfrac{d}{dx}[c\cdot f(x)] = c\cdot f'(x)$ | A numeric coefficient on a function. Differentiate the function and keep the coefficient out front. | | Sum and difference | $\dfrac{d}{dx}[f \pm g] = f' \pm g'$ | A function written as terms joined by $+$ or $-$. Handle each term separately. | | Product | $\dfrac{d}{dx}[fg] = f'g + fg'$ | Two nonconstant functions multiplied. The two terms are added, so their order is free. | | Quotient | $\dfrac{d}{dx}\left[\dfrac{f}{g}\right] = \dfrac{g f' - f g'}{g^{2}}$ | One function divided by another when the bottom is not a single power. Order is fixed: the $g f'$ term comes first. | | Chain | $\dfrac{d}{dx}\big[f(g(x))\big] = f'(g(x)) \cdot g'(x)$ | A composition, one function inside another. Multiply the outside derivative by the inside derivative. | NOTE (The two most common slips): The product rule is $f'g + fg'$, never $f'g'$. The quotient rule subtracts in a fixed order, and reversing the two numerator terms flips the sign of the whole answer. ### The six trigonometric derivatives (Topics 2.7 and 2.10) Sine and cosine are the two you memorize outright (Topic 2.7). The other four come from rewriting tangent, cotangent, secant, and cosecant with identities and differentiating (Topic 2.10). All six formulas below assume the angle is in radians. | Function | Derivative | Why / when | | --- | --- | --- | | $\sin x$ | $\cos x$ | Memorize directly; the anchor for the rest. | | $\cos x$ | $-\sin x$ | A co-function, so its derivative carries a minus sign. | | $\tan x$ | $\sec^{2} x$ | From the quotient rule on $\sin x / \cos x$. | | $\cot x$ | $-\csc^{2} x$ | Co-function of tangent; note the minus sign. | | $\sec x$ | $\sec x \tan x$ | From the chain rule on $(\cos x)^{-1}$. | | $\csc x$ | $-\csc x \cot x$ | Co-function of secant; carries a minus sign. | NOTE (Two patterns that cut the memorizing in half): Every co-function (cosine, cotangent, cosecant) has a derivative that carries a minus sign. And when the angle is a function of $x$ rather than a bare $x$, the chain rule tacks on the inside derivative: $\dfrac{d}{dx}[\sin u] = \cos u \cdot u'$. ### Inverse trig, exponential, and log derivatives (Topics 2.7 and 3.4) The exponential and log rules for $e^{x}$ and $\ln x$ are Topic 2.7; the general bases $a^{x}$ and $\log_a x$ come from them by rewriting. Notice how $\ln a$ shows up as an extra factor whenever the base is not $e$. | Function | Derivative | Why / when | | --- | --- | --- | | $e^{x}$ | $e^{x}$ | Equal to its own derivative: every solution of $f' = f$ is $Ce^{x}$. | | $a^{x}$ | $a^{x} \ln a$ | General base: the extra factor is $\ln a$. When $a = e$, $\ln e = 1$ recovers $e^{x}$. | | $\ln x$ | $\dfrac{1}{x}$ | For $x > 0$. Turns a logarithm into a simple power. | | $\log_a x$ | $\dfrac{1}{x \ln a}$ | Change of base puts $\ln a$ in the denominator. | The inverse trig derivatives (Topic 3.4) come from the inverse-function rule and the Pythagorean identities. AB and BC most often test $\arcsin$, $\arccos$, and $\arctan$; the other three are listed for completeness. The co-function pattern repeats: each co-inverse has the same size derivative as its partner but the opposite sign. | Function | Derivative | Why / when | | --- | --- | --- | | $\arcsin x$ | $\dfrac{1}{\sqrt{1-x^{2}}}$ | Domain $|x| < 1$. The inverse trig derivative AP uses most. | | $\arccos x$ | $-\dfrac{1}{\sqrt{1-x^{2}}}$ | Same size as $\arcsin$, opposite sign (co-function). | | $\arctan x$ | $\dfrac{1}{1+x^{2}}$ | Defined for all $x$; no radical, no sign trap. | | $\operatorname{arccot} x$ | $-\dfrac{1}{1+x^{2}}$ | Co-function of $\arctan$; opposite sign. | | $\operatorname{arcsec} x$ | $\dfrac{1}{|x|\sqrt{x^{2}-1}}$ | Domain $|x| > 1$; keep the absolute value: dropping it flips the sign for $x < -1$. | | $\operatorname{arccsc} x$ | $-\dfrac{1}{|x|\sqrt{x^{2}-1}}$ | Co-function of $\operatorname{arcsec}$; opposite sign. | NOTE (There is no formula sheet): The AP Calculus AB and BC exams provide no reference sheet, so every rule on this page has to be memorized or rebuilt quickly. With a chain rule, each inside function $u$ contributes a factor of $u'$: for example $\dfrac{d}{dx}[\arctan u] = \dfrac{u'}{1+u^{2}}$. ## Common Integrals: The Complete Table URL: https://www.calclearn.org/reference/common-integrals Last updated: 2026-07-22 Answer first: Every AP antiderivative comes from reversing a derivative you know: the power rule (add one to the exponent, divide), 1/x gives ln|x|, e^x stays e^x, a^x gives a^x/ln a, plus the six trig and two inverse-trig forms. All are AB and BC; BC adds techniques, not new forms. Always write + C. ### How to read this table: reverse a derivative you know An antiderivative is a derivative rule run backward, so the recognition skill this whole reference trains is one question: which derivative would have produced this integrand? Topic 6.8 supplies the basic rules, and every form on this page is one you can rebuild by differentiating the answer and checking that you land back on the integrand. Nothing in the core table below is BC only. Each row is on both AB and BC, and the section at the end says exactly what BC adds. | Integral | Antiderivative | When to reach for it | | --- | --- | --- | | $\int x^n\,dx$ | $\dfrac{x^{n+1}}{n+1}+C,\ n\neq -1$ | Any power of $x$: whole, fractional, or negative. Add one to the exponent, divide by the new exponent. Rewrite roots as $x^{1/2}$ and $\frac{1}{x^3}$ as $x^{-3}$ first. | | $\int \dfrac{1}{x}\,dx$ | $\ln|x|+C$ | The $n=-1$ case the power rule cannot touch (dividing by $n+1=0$ fails). The absolute value keeps it valid where $x<0$. | | $\int e^x\,dx$ | $e^x+C$ | $e^x$ is its own antiderivative. If the exponent is $kx$ instead of $x$, divide by $k$ (see the linear-inside table). | | $\int a^x\,dx$ | $\dfrac{a^x}{\ln a}+C$ | Base $a>0,\ a\neq 1$. Reverses $(a^x)'=a^x\ln a$, so you divide by $\ln a$. Setting $a=e$ recovers the row above, since $\ln e=1$. | NOTE (The + C is not optional): An indefinite integral names a whole family of functions that differ by a constant. Drop the $+C$ and you have named one member, not the antiderivative. On a free-response question that gives an initial condition, the missing $+C$ is exactly where the particular-solution point disappears. Definite integrals need no $+C$: it cancels in $F(b)-F(a)$. ### Trigonometric antiderivatives These six are the derivative rules of Topics 2.7 and 2.10 read backward, nothing more. Match the integrand to the derivative that produces it and copy the answer. The only recurring error is sign. Three of the six carry a minus: they are exactly the ones whose antiderivative is a cofunction ($-\cos$, $-\cot$, $-\csc$), because the cofunctions $\cos$, $\cot$, and $\csc$ are the ones whose derivatives carry the minus sign. | Integral | Antiderivative | Reverses / watch for | | --- | --- | --- | | $\int \sin x\,dx$ | $-\cos x + C$ | Reverses $(\cos x)'=-\sin x$. The leading minus sign is the single most common error on this list. | | $\int \cos x\,dx$ | $\sin x + C$ | Reverses $(\sin x)'=\cos x$. No sign flip. | | $\int \sec^2 x\,dx$ | $\tan x + C$ | Reverses $(\tan x)'=\sec^2 x$. Spot the squared secant. | | $\int \csc^2 x\,dx$ | $-\cot x + C$ | Reverses $(\cot x)'=-\csc^2 x$. Carries a minus. | | $\int \sec x\tan x\,dx$ | $\sec x + C$ | Reverses $(\sec x)'=\sec x\tan x$. The product is the tell. | | $\int \csc x\cot x\,dx$ | $-\csc x + C$ | Reverses $(\csc x)'=-\csc x\cot x$. Carries a minus. | Tangent and cotangent are not on the basic list because you do not get them by reversing a single derivative; you get them from a u-substitution (Topic 6.9). They still show up often enough to keep nearby. | Integral | Antiderivative | How you actually get it | | --- | --- | --- | | $\int \tan x\,dx$ | $-\ln|\cos x| + C = \ln|\sec x| + C$ | Let $u=\cos x$, so $du=-\sin x\,dx$: it becomes $\int \frac{-1}{u}\,du=-\ln|u|+C=-\ln|\cos x|+C$. | | $\int \cot x\,dx$ | $\ln|\sin x| + C$ | Let $u=\sin x$, so $du=\cos x\,dx$: it becomes $\int \frac{1}{u}\,du=\ln|u|+C=\ln|\sin x|+C$. | NOTE (Do not memorize the integral of secant): $\int \sec x\,dx=\ln|\sec x+\tan x|+C$ is a standard formula in a college course, but it is outside the AP forms and is not expected on the exam. If a problem seems to demand it, re-read the integrand: the intended path is almost always a u-substitution or one of the rows above. ### Inverse-trigonometric antiderivatives These are where a fraction with no obvious substitution turns into an inverse trig function. Recognizing the pattern is the entire skill, because the integrand does not look like anything until you match the shape of the denominator. Topic 6.10 (completing the square) exists largely to push a completed-square denominator into the arctan form, and it is on both AB and BC. | Integral | Antiderivative | Pattern to recognize | | --- | --- | --- | | $\int \dfrac{1}{\sqrt{1-x^2}}\,dx$ | $\arcsin x + C$ | A root over $\text{constant}-x^2$ with a plain numerator. A square root in the denominator points to $\arcsin$. | | $\int \dfrac{1}{1+x^2}\,dx$ | $\arctan x + C$ | $\text{constant}+x^2$ in the denominator, no root. That shape points to $\arctan$. | | $\int \dfrac{1}{\sqrt{a^2-x^2}}\,dx$ | $\arcsin\dfrac{x}{a} + C$ | The general radius-$a$ version. Same shape, with $a^2$ in place of $1$; the answer divides the inside by $a$, giving $\arcsin\dfrac{x}{a}$. | | $\int \dfrac{1}{a^2+x^2}\,dx$ | $\dfrac{1}{a}\arctan\dfrac{x}{a} + C$ | The general $a$ version. Note the extra $\frac{1}{a}$ out front, which the $\arcsin$ form does not have. | NOTE (Which inverse-trig form is it?): A square root in the denominator means $\arcsin$; no root means $\arctan$. If the denominator is a quadratic that is not yet in $a^2\pm x^2$ shape, complete the square first (Topic 6.10), then read off $a$. The arcsecant form $\int \frac{1}{x\sqrt{x^2-1}}\,dx=\text{arcsec}|x|+C$ is essentially never tested on the AP exam; the two patterns above cover what you need. ### Constant multiples, a linear inside, and what BC adds Two adjustments extend every row above without new memorizing. Constant multiples pull straight out front: $\int k\,f(x)\,dx = k\int f(x)\,dx$. And when the inside is linear, of the form $ax+b$, a single u-substitution shows the only change to the answer is dividing by $a$. Knowing these shortcut versions cold saves you a full written substitution on the exam. | Integral (linear inside) | Antiderivative | Why the extra constant appears | | --- | --- | --- | | $\int e^{kx}\,dx$ | $\dfrac{1}{k}e^{kx}+C$ | $u=kx$ gives $du=k\,dx$, so the $\frac{1}{k}$ is that constant coming back out. | | $\int \cos(kx)\,dx$ | $\dfrac{1}{k}\sin(kx)+C$ | Same $u=kx$. The sign follows the plain $\int \cos x\,dx$ rule. | | $\int \sin(kx)\,dx$ | $-\dfrac{1}{k}\cos(kx)+C$ | Same $u=kx$. Keep the minus from $\int \sin x\,dx$. | | $\int (ax+b)^n\,dx$ | $\dfrac{(ax+b)^{n+1}}{a(n+1)}+C,\ n\neq -1$ | Power rule with an extra $\frac{1}{a}$ from $u=ax+b$. | | $\int \dfrac{1}{ax+b}\,dx$ | $\dfrac{1}{a}\ln|ax+b|+C$ | The $n=-1$ case of the row above. Still a log, still divided by $a$. | NOTE (BC note: no new basic forms, just new ways in): Every form on this page is on both AB and BC. BC does not add antiderivative formulas; it adds techniques for rewriting a hard integral until one of these rows applies: integration by parts (Topic 6.11), linear partial fractions (Topic 6.12), and improper integrals (Topic 6.13). Partial fractions, for example, only turns a rational function into a sum of $\frac{1}{ax+b}$ pieces, each of which uses the log row above. The table you memorize is identical for both courses. ## Limit Laws and Indeterminate Forms URL: https://www.calclearn.org/reference/limit-laws-and-indeterminate-forms Last updated: 2026-07-22 Answer first: Only 0/0 and infinity/infinity qualify for L'Hospital's Rule; forms like (nonzero)/0 or infinity+infinity only look indeterminate. Split across sums, products, and quotients when each piece has a real limit; for quotients, the denominator's limit must be nonzero. Memorize sin(x)/x = 1 and (1-cos x)/x = 0 near x = 0. ### The limit laws NOTE (Always try direct substitution first): For any polynomial, rational, power, exponential, logarithmic, or trig function, $\lim_{x \to c} f(x) = f(c)$ at every point in its domain (Topic 1.12). Plug in $c$. If you get a real number, that is the answer. The laws below justify why substitution works and tell you what to do when it fails. Each law assumes the individual limits exist as real numbers: let $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$ (Topic 1.5). | Law | Rule | When to use it | | --- | --- | --- | | Sum and difference | $\lim_{x \to c}\left[f(x) \pm g(x)\right] = L \pm M$ | Break any expression joined by $+$ or $-$ into separate limits. | | Constant multiple | $\lim_{x \to c}\left[k \cdot f(x)\right] = k \cdot L$ | Pull a numerical coefficient outside the limit. | | Product | $\lim_{x \to c}\left[f(x) \cdot g(x)\right] = L \cdot M$ | Split a product into factors you can evaluate on their own. | | Quotient | $\lim_{x \to c} \dfrac{f(x)}{g(x)} = \dfrac{L}{M}, \ M \neq 0$ | Valid only when the denominator limit is nonzero. If $M = 0$, stop: you have $0/0$ or an infinite limit. | | Power | $\lim_{x \to c}\left[f(x)\right]^{n} = L^{n}$ | Valid for $n$ a positive integer. Evaluate the inside first, then raise to the power. | | Root | $\lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{L}$ | Needs $L \ge 0$ when $n$ is even. | | Constant | $\lim_{x \to c} k = k$ | A constant function ignores $x$. | | Identity | $\lim_{x \to c} x = c$ | The building block behind direct substitution. | | Composite | $\lim_{x \to c} f\left(g(x)\right) = f(M)$ | Valid when $f$ is continuous at $M$: take the inside limit, then apply the outer function. | ### Indeterminate vs forms that only look indeterminate When direct substitution returns one of exactly two patterns, the limit is genuinely indeterminate and L'Hospital's Rule (Topic 4.7) is on the table. Everything else that looks alarming actually settles to a definite answer. | Substitution gives | Status | What to do | | --- | --- | --- | | $\dfrac{0}{0}$ | Indeterminate | First try factoring, a conjugate, or a special trig limit. If none apply, use L'Hospital's Rule (Topic 4.7). | | $\dfrac{\infty}{\infty}$ | Indeterminate | Compare growth rates or apply L'Hospital's Rule. For a ratio of polynomials, the ratio of leading terms also gives the answer. | | Substitution gives | Actual result | Why | | --- | --- | --- | | $\dfrac{k}{0}, \ k \neq 0$ | $+\infty$, $-\infty$, or DNE | A vertical asymptote. Check the sign of the denominator from each side (Topic 1.14). | | $\dfrac{0}{k}, \ k \neq 0$ | $0$ | A vanishing numerator over a fixed nonzero number is $0$. | | $\dfrac{k}{\infty}$ | $0$ | A fixed number divided by something unbounded shrinks to $0$. | | $\infty + \infty$ | $+\infty$ | Two quantities both growing without bound add to $\infty$. | | $k \cdot \infty, \ k \neq 0$ | $\pm\infty$ | A nonzero constant times an unbounded quantity stays unbounded; the sign of $k$ fixes the sign. | | $\infty - k$ | $+\infty$ | Subtracting a finite number never tames an unbounded one. | NOTE (Genuinely indeterminate, but off the exam): The forms $\infty - \infty$, $0 \cdot \infty$, $1^{\infty}$, $0^{0}$, and $\infty^{0}$ are indeterminate too, but the CED excludes them from both AB and BC (Topic 4.7 exclusion statement). On the exam, L'Hospital's Rule always starts from $0/0$ or $\infty/\infty$. ### The two special trig limits Both special limits come from the Squeeze Theorem (Topic 1.8). They turn a $0/0$ trig form into a number without L'Hospital, but only when the angle inside the trig function matches the variable in the denominator. | Limit | Value | How to spot and use it | | --- | --- | --- | | $\lim_{x \to 0} \dfrac{\sin x}{x}$ | $1$ | The argument of $\sin$ must equal the denominator, and both must approach $0$. Force a match by multiplying and dividing. | | $\lim_{x \to 0} \dfrac{1 - \cos x}{x}$ | $0$ | Same matching rule. The value is $0$ whichever way you write the numerator, since both $1 - \cos x$ and $\cos x - 1$ vanish faster than $x$. | | $\lim_{x \to 0} \dfrac{\tan x}{x}$ | $1$ | Follows from $\dfrac{\sin x}{x} \cdot \dfrac{1}{\cos x}$. Handy, but not required to memorize. | Worked example: evaluate $\lim_{x \to 0} \dfrac{\sin 5x}{x}$. Direct substitution gives $0/0$, and the argument $5x$ does not match the denominator $x$. Multiply the numerator and denominator by $5$ so the denominator matches the angle: $$\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} \frac{5 \sin 5x}{5x} = 5 \lim_{x \to 0} \frac{\sin 5x}{5x}$$ As $x \to 0$, the angle $5x \to 0$ as well, so $\dfrac{\sin 5x}{5x} \to 1$. Therefore the limit is $5 \cdot 1 = 5$. $$\lim_{x \to 0} \frac{\sin 5x}{x} = 5$$ ### Choosing the right tool Method selection for limits is driven almost entirely by what direct substitution produces. Read off the form, then pick the matching technique. | Substitution produces | Read it as | Technique | | --- | --- | --- | | A real number | Continuous point | Done: that number is the limit (Topic 1.5). | | $0/0$ with polynomials or radicals | Removable form | Factor and cancel, or multiply by the conjugate, then re-substitute (Topic 1.6). | | $0/0$ with $\sin x$ or $1 - \cos x$ near $0$ | Special trig limit | Match the argument to $\dfrac{\sin x}{x}$ or $\dfrac{1 - \cos x}{x}$ (Topic 1.8). | | $0/0$ or $\infty/\infty$ that will not simplify | True indeterminate | L'Hospital's Rule: differentiate the top and bottom separately (Topic 4.7). | | $k/0$ with $k \neq 0$ | Infinite limit | Find the sign from each side; the answer is $+\infty$, $-\infty$, or DNE (Topic 1.14). | | Any determinate form from the table above | Not indeterminate | Read the value straight off; no special work needed. | $$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$$ NOTE (The L'Hospital trap): The rule above differentiates the numerator and denominator separately: it is not the quotient rule. Re-check the form after each pass, because one application can turn $0/0$ into a real number (you are done) or into a fresh indeterminate form (apply it again). ## Series Convergence Tests: The Complete Table (BC) URL: https://www.calclearn.org/reference/convergence-tests Last updated: 2026-07-22 Answer first: Eight tests cover AP Calculus BC series (CED 10.1 to 10.9): nth-term, geometric, p-series, integral, direct comparison, limit comparison, alternating, and ratio. Match the shape of the terms to the test, run the nth-term check first, and remember it proves divergence only, never convergence. ### The eight tests at a glance These eight are the only convergence tools the AP Calculus BC exam assesses. The CED (the College Board's Course and Exam Description) Topic 10.8 exclusion statement names the six formal tests it assesses (nth-term, integral, comparison, limit comparison, alternating series, and ratio); geometric series (10.2) and p-series (10.5) are the two known series you add to complete the toolkit. Read the table left to right: match the shape of the terms in $\sum a_n$ to a test, confirm what that test can actually conclude, then read the trap that costs the most points. Series live in Unit 10, worth 15 to 20 percent of the exam. | Test | Applies to | Concludes | Watch out | | --- | --- | --- | --- | | nth-Term Test (10.3) | Any series, as the opening check | If $\lim_{n \to \infty} a_n \ne 0$ (or the limit does not exist), diverges. If the limit is $0$, inconclusive. | Never proves convergence. $a_n \to 0$ is necessary, not sufficient: the harmonic series obeys it and still diverges. | | Geometric Series (10.2) | A constant ratio $r$ between successive terms, $\sum a r^n$ | Converges to $\frac{a}{1-r}$ if $|r| < 1$; diverges if $|r| \ge 1$. | Before using $\frac{a}{1-r}$, check the starting index: $a$ is the first term actually written out. | | p-Series (10.5) | Terms of the form $\frac{1}{n^p}$ | Converges if $p > 1$; diverges if $p \le 1$. | $p = 1$ is the harmonic series and diverges; $p = \frac{1}{2}$ (a $\frac{1}{\sqrt{n}}$ term) also diverges. | | Integral Test (10.4) | $a_n = f(n)$ with $f$ positive, continuous, and decreasing for $x \ge k$ | $\sum a_n$ and $\int_k^\infty f(x)\,dx$ converge or diverge together. | The integral's value is not the sum. You must be able to antidifferentiate $f$ for this to be usable. | | Direct Comparison (10.6) | Positive terms you can bound by a known series | If $0 \le a_n \le b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges; if $0 \le b_n \le a_n$ and $\sum b_n$ diverges, then $\sum a_n$ diverges. | Direction is everything: bound above by a convergent series to prove convergence, below by a divergent one to prove divergence. Any other pairing proves nothing. | | Limit Comparison (10.6) | Rational or algebraic terms close to a p-series | With $b_n > 0$, if $L = \lim_{n \to \infty} \frac{a_n}{b_n}$ and $0 < L < \infty$, both series do the same thing. | The limit must be finite and positive. Pick $b_n$ from the leading powers of $a_n$. | | Alternating Series Test (10.7) | Alternating signs, $\sum (-1)^n b_n$ with $b_n > 0$ | Converges if $b_n$ is decreasing and $\lim_{n \to \infty} b_n = 0$. | Both conditions are required, and the rubric expects you to state that $b_n$ decreases. Proves convergence only, never divergence. | | Ratio Test (10.8) | Factorials, exponentials, or $n$th powers | $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$: $L < 1$ converges absolutely, $L > 1$ diverges, $L = 1$ inconclusive. | $L = 1$ on every p-series and rational term. When it happens, switch to a comparison test. | NOTE (Run the nth-term test first): It is one limit and it can finish the problem: if $\lim_{n \to \infty} a_n \ne 0$, the series diverges immediately. If the limit is $0$, you have learned nothing yet, so keep going. Writing "the terms go to zero, so it converges" is the single most common lost point on series. ### The two series you memorize Two series are pure recall on the exam: you never derive them, you recognize them. Every comparison you set up eventually leans on one of these, so memorize the cutoffs cold. | Series | Form | Converges when | Key fact | | --- | --- | --- | --- | | Geometric Series (10.2) | $\sum_{n=0}^{\infty} a r^n$ | $|r| < 1$ | Sum equals $\frac{a}{1-r}$, the only series in this list with a general closed-form sum formula you can evaluate for any $a$ and $r$. | | p-Series (10.5) | $\sum_{n=1}^{\infty} \frac{1}{n^p}$ | $p > 1$ | No exam-required closed sum; it is the benchmark almost every comparison targets. | | Harmonic Series (10.5) | $\sum_{n=1}^{\infty} \frac{1}{n}$ | Never (diverges) | The $p = 1$ case: terms shrink to $0$, yet the series diverges. The canonical counterexample. | | Alternating Harmonic Series (10.5) | $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$ | Always (converges) | Conditionally convergent: it converges to $\ln 2$, but drop the signs and it diverges. | $$\sum_{n=0}^{\infty} a r^n = \frac{a}{1-r}, \quad |r| < 1$$ ### Recognition cues: reading the terms Method selection is the whole game: most lost time on series comes from starting the wrong test. Each pattern in the terms points to one test. Find the row that matches what $a_n$ looks like, then run it. | What the terms look like | Reach for | Why it works | | --- | --- | --- | | A constant multiplier between successive terms | Geometric Series (10.2) | Divide any term by the previous one; if it is always the same $r$, convergence reads straight off $|r|$. | | $\frac{1}{n^p}$ | p-Series (10.5) | Convergence is decided by a single number, $p$, with the cutoff at $1$. | | A factorial $n!$, an exponential $c^n$, or an $n$th power | Ratio Test (10.8) | Forming $\frac{a_{n+1}}{a_n}$ cancels the growing factor and leaves a clean limit. | | A rational or algebraic function of $n$ | Limit Comparison (10.6) | Strip to the leading powers to find the p-series it behaves like, then confirm with the limit. | | Signs that flip through $(-1)^n$ or $(-1)^{n-1}$ | Alternating Series Test (10.7) | Only two conditions to check; a series that fails other tests may still converge here. | | $a_n = f(n)$ with $f$ positive, decreasing, and easy to antidifferentiate | Integral Test (10.4) | The improper integral settles it. Natural for $\frac{1}{n \ln n}$-type terms with a logarithm. | | Terms trapped between a known series and $0$ | Direct Comparison (10.6) | When the bounding inequality is obvious, no limit is needed. | | Anything, as the opening move | nth-Term Test (10.3) | One limit can end the problem before any real work begins. | ### Absolute versus conditional convergence (10.9) Once a series converges, Topic 10.9 asks a second question: is the convergence absolute or conditional? Test the series of absolute values $\sum |a_n|$ and match your result to the table. | Situation | Classification | What it means | | --- | --- | --- | | $\sum |a_n|$ converges | Absolutely convergent | Convergence is guaranteed (absolute convergence implies convergence), and rearranging the terms keeps the same sum. | | $\sum a_n$ converges but $\sum |a_n|$ diverges | Conditionally convergent | The series survives only through sign cancellation; the alternating harmonic series is the model case. | | $\sum a_n$ diverges | Divergent | Neither classification applies. | NOTE (A shortcut worth memorizing): Before running the alternating series test, try the ratio test on $|a_n|$. If it returns $L < 1$, you already have absolute convergence, which implies convergence, and you can skip the alternating series test entirely. If it returns $L > 1$, the terms do not go to zero, so the series diverges by the nth-term test. If it returns $L = 1$, fall back to the alternating series test. ## Common Maclaurin Series Table (BC) URL: https://www.calclearn.org/reference/common-maclaurin-series Last updated: 2026-07-22 Answer first: The four Maclaurin series to memorize are e^x = sum of x^n/n! (all x), sin x = sum of (-1)^n x^(2n+1)/(2n+1)! (all x), cos x = sum of (-1)^n x^(2n)/(2n)! (all x), and 1/(1-x) = sum of x^n (only for |x| < 1). Two common extensions, ln(1+x) and arctan x, hold for -1 < x <= 1 and |x| <= 1. ### The four series the CED names All of Unit 10 is BC-only, and these four Maclaurin series are the ones the College Board builds on directly in Topic 10.14: $e^x$, $\sin x$, and $\cos x$ are named as the foundation for constructing other series (LIM-8.F.2), and $\frac{1}{1-x}$ is the geometric building block (LIM-8.F.1). Memorize these four cold. Almost every other series on the BC exam is reached by reshaping one of them. | Function | Maclaurin series (general term) | Converges for | When to reach for it | | --- | --- | --- | --- | | $e^x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ | all real $x$ | Any $e^{(\text{expression})}$: substitute the exponent for $x$. Converges everywhere, so there is no interval to check. | | $\sin x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ | all real $x$ | Odd powers only, matching that $\sin$ is an odd function. Use for $\sin$ of any expression; safe to substitute freely. | | $\cos x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ | all real $x$ | Even powers only, matching that $\cos$ is even. It is the term-by-term derivative of the $\sin x$ series, so you can recover one from the other. | | $\frac{1}{1-x}$ | $\displaystyle\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$ | $|x| < 1$ | The geometric series with ratio $x$ (LIM-8.F.1). Base for $\frac{1}{1+x}$, $\frac{1}{1-x^2}$, and similar rationals. Diverges once $|x| \ge 1$, so always carry the interval. | The three that converge for all $x$ ($e^x$, $\sin x$, $\cos x$) are the ones you can substitute into without a second thought: any expression can replace $x$ and the result still converges everywhere. The geometric series is the exception. It collapses the moment $|x| \ge 1$, so any substitution into it comes with an interval condition attached. ### Two extensions worth memorizing These two are not in the CED (Course and Exam Description) essential knowledge, so treat them as commonly-taught extras rather than required facts. You can derive each by integrating a geometric-type series term by term (Topic 10.15), but memorizing them saves time. They are labeled here as extensions so you know they sit outside the core four above. | Function (extension) | Maclaurin series (general term) | Converges for | Where it comes from | | --- | --- | --- | --- | | $\ln(1+x)$ | $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ | $-1 < x \le 1$ | Integrate the $\frac{1}{1+x}$ series term by term. The sum starts at $n = 1$: there is no constant term because $\ln 1 = 0$. | | $\arctan x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$ | $-1 \le x \le 1$ | Integrate the $\frac{1}{1+x^2}$ series term by term. Odd powers like $\sin x$, but the denominators are $2n+1$, not factorials. | Both extensions come from integrating a known series, which is why their intervals pick up an endpoint the plain geometric series does not have. Term-by-term integration keeps the radius of convergence the same (LIM-8.D.6) but can change what happens at the endpoints, so $\ln(1+x)$ converges at $x = 1$ and $\arctan x$ converges at both $x = \pm 1$. An interval question still needs each endpoint tested on its own (Topic 10.13). ### Choosing which series to start from Method selection is the real skill. Before computing anything, look at $f(x)$ and decide which base series it is a disguised version of. The derivative formula $\frac{f^{(n)}(a)}{n!}$ (Topic 10.11) always works but is slow, so save it for functions that match nothing on the list. This table maps a function's shape to the series and the single operation that builds it fastest. | If $f(x)$ looks like | Start from | Do this | Example | | --- | --- | --- | --- | | A standard function with $x$ replaced by an expression: $\cos(3x)$, $e^{-x^2}$ | the matching base series | substitute the expression in place of $x$ | $e^{-x^2} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ | | A base series times a power of $x$: $x^2 \sin x$ | the base series | multiply every term by that power | $x^2 \sin x = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+3}}{(2n+1)!}$ | | A rational form $\frac{1}{1 \pm ax}$ or $\frac{1}{1 \pm ax^2}$ | $\frac{1}{1-x}$ | substitute, factoring out constants first | $\frac{1}{1+x^2} = \displaystyle\sum_{n=0}^{\infty} (-1)^n x^{2n}$ | | A function whose derivative is a known series: $\ln(1+x)$, $\arctan x$ | the derivative's series (e.g. $\frac{1}{1+x}$ or $\frac{1}{1+x^2}$) | integrate term by term, then fix the constant | $\int \frac{1}{1+x^2}\,dx = \arctan x$ | | Nothing on the list, or a center $a \ne 0$ | the derivative formula | compute $\frac{f^{(n)}(a)}{n!}$ and attach to $(x-a)^n$ | $\ln x$ centered at $a = 1$ | You can chain these operations: for $x^2 \cos(2x)$, substitute $2x$ into the cosine series, then multiply every term by $x^2$, giving $x^2 \cos(2x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n+2}}{(2n)!}$. The one case with no shortcut is a series centered at $a \ne 0$, such as $\ln x$ at $x = 1$; there you compute derivatives at $a$ and attach each coefficient to $(x-a)^n$, not $x^n$. The Taylor and Maclaurin series guide works that example in full. ### Watch-outs when using the table A handful of slips account for most lost points once you start pulling series from this table. - Factorials belong to $e^x$, $\sin x$, and $\cos x$ only. The geometric series and its integrals ($\frac{1}{1-x}$, $\ln(1+x)$, $\arctan x$) have plain integer denominators or none at all. - Signs travel through a substitution: $(-x^2)^n = (-1)^n x^{2n}$, so replacing $x$ with $-x^2$ changes both the sign pattern and the powers. - Intervals do not carry over unchanged. Substituting $u = g(x)$ into a series valid for $|u| < 1$ gives the condition $|g(x)| < 1$, which you still have to solve for $x$. - Mind the starting index. The $\ln(1+x)$ series starts at $n = 1$ (its first term is $x$), while $e^x$, $\cos x$, and $\frac{1}{1-x}$ start at $n = 0$. ## Trig Identities You Actually Need for Calculus URL: https://www.calclearn.org/reference/trig-identities-for-calculus Last updated: 2026-07-22 Answer first: You need three families for AP Calculus: the Pythagorean identities (sin^2 x + cos^2 x = 1 and its two divided-down forms), the double-angle formulas for cosine that power-reduce sin^2 x and cos^2 x, and the sum formulas. Each one exists to rewrite a trig expression you cannot antidifferentiate into one you can. ### When you reach for a trig identity You pull out a trig identity when an expression is trigonometric and no basic antiderivative or single u-substitution fits it as written. The identity's only job is to rewrite that trig into a shape one of your known rules can finish. This is a rewrite-then-recognize task, and in AP it lives mostly in Topic 6.9 (integrating using substitution) and the derivative rules of Topics 2.7 and 2.10 that those integrals reverse. NOTE (The one question that decides): Does a basic antiderivative or a clean u-substitution already stare back at you? If yes, you need no identity. If instead you see an even power of sine or cosine, a squared tangent or cotangent, or a product like $\sin x\cos x$, the identity that unlocks it is on this page. There is no AP formula sheet, so you rebuild these from memory, but the tables below show you only need to memorize a handful and derive the rest. | When the integrand looks like | Rewrite it with | Because it becomes | | --- | --- | --- | | $\int \sin^2 x\,dx$ or $\int \cos^2 x\,dx$ | A power-reduction form of $\cos 2x$ | A squared trig function has no basic antiderivative; the linear $\cos 2x$ that replaces it does. | | $\int \tan^2 x\,dx$ | $1+\tan^2 x=\sec^2 x$ | $\tan^2 x=\sec^2 x-1$, and $\int \sec^2 x\,dx=\tan x$. | | $\int \cot^2 x\,dx$ | $1+\cot^2 x=\csc^2 x$ | $\cot^2 x=\csc^2 x-1$, and $\int \csc^2 x\,dx=-\cot x$. | | An odd power, like $\int \sin^3 x\,dx$ | $\sin^2 x=1-\cos^2 x$ | Peels off one $\sin x$ for $du$ and leaves cosines, so $u=\cos x$ finishes it (Topic 6.9). | | $\int 2\sin x\cos x\,dx$ | $\sin 2x=2\sin x\cos x$ | Collapses the product into a single sine you integrate in one step. | | $\sqrt{\sin^2\theta+\cos^2\theta}$ inside an integrand | $\sin^2 x+\cos^2 x=1$ | The radical simplifies to a constant, common in BC arc-length and parametric-speed setups. | ### The three Pythagorean identities All three are one identity seen from three angles. Start from $\sin^2 x+\cos^2 x=1$. Divide every term by $\cos^2 x$ and you get the tangent-secant form; divide every term by $\sin^2 x$ and you get the cotangent-cosecant form. You do not memorize three separate facts, you memorize one and divide. | Identity | How you get it | Its calculus job | | --- | --- | --- | | $\sin^2 x+\cos^2 x=1$ | Memorize outright (the unit circle) | Swap $\sin^2 x$ for $1-\cos^2 x$ (or the reverse) to free one factor for u-substitution, and collapse radicals in arc-length and parametric-speed integrands. | | $1+\tan^2 x=\sec^2 x$ | Divide the first identity by $\cos^2 x$ | Turns $\int \tan^2 x\,dx$ into $\int(\sec^2 x-1)\,dx$, because $\sec^2 x$ is the member you can antidifferentiate. | | $1+\cot^2 x=\csc^2 x$ | Divide the first identity by $\sin^2 x$ | The same move for cotangent: $\int \cot^2 x\,dx=\int(\csc^2 x-1)\,dx$. | The tangent case is the model worked example. You cannot integrate $\tan^2 x$ directly, but the identity trades it for something you can: $$\int \tan^2 x\,dx = \int (\sec^2 x - 1)\,dx = \int \sec^2 x\,dx - \int 1\,dx = \tan x - x + C$$ NOTE (Why sec squared and csc squared are the targets): You always convert toward $\sec^2 x$ and $\csc^2 x$, never away from them, because $\sec^2 x$ is exactly the derivative of $\tan x$ and $\csc^2 x$ is exactly the derivative of $-\cot x$: in prime notation $(\tan x)' = \sec^2 x$ and $(\cot x)' = -\csc^2 x$ (Topic 2.10). They are the only members of this family with a one-step antiderivative, $\tan x$ and $-\cot x$ respectively, so the identity exists to move your integrand onto that landing spot. ### Double-angle formulas and power reduction The double-angle formulas matter in calculus for one reason above all others: they are how you integrate an even power of sine or cosine. A squared trig function is not on your basic antiderivative list, so you trade it for a first power of $\cos 2x$, which is. Cosine has three equivalent forms, and the two rewritten forms are the ones that do the work. | Formula | Which form, and when | | --- | --- | | $\sin 2x = 2\sin x\cos x$ | Read right-to-left to collapse a $2\sin x\cos x$ product into one term; read left-to-right when a $\sin 2x$ needs breaking apart. | | $\cos 2x = \cos^2 x - \sin^2 x$ | The base form. The next two rows rewrite it with $\sin^2 x+\cos^2 x=1$ and are the ones you actually integrate with. | | $\cos 2x = 2\cos^2 x - 1$ | Solve this for $\cos^2 x$ to get the power-reduction formula for cosine. | | $\cos 2x = 1 - 2\sin^2 x$ | Solve this for $\sin^2 x$ to get the power-reduction formula for sine. | Solving those last two forms for the squared term gives the payoff, the two power-reduction formulas you reach for whenever an even power of sine or cosine stands alone in an integral: | Power-reduction form | Comes from solving | Use it to integrate | | --- | --- | --- | | $\cos^2 x = \frac{1+\cos 2x}{2}$ | $\cos 2x = 2\cos^2 x - 1$ | $\int \cos^2 x\,dx$: the right side is a constant plus $\cos 2x$, both one-step integrals. | | $\sin^2 x = \frac{1-\cos 2x}{2}$ | $\cos 2x = 1 - 2\sin^2 x$ | $\int \sin^2 x\,dx$: the same, carrying a minus sign. | Here is the sine case worked all the way through, using $\int \cos 2x\,dx=\frac{1}{2}\sin 2x$ for the linear inside: $$\begin{aligned}\int \sin^2 x\,dx &= \int \frac{1-\cos 2x}{2}\,dx = \frac{1}{2}\int (1-\cos 2x)\,dx \\ &= \frac{1}{2}\left(x - \frac{1}{2}\sin 2x\right)+C = \frac{x}{2} - \frac{\sin 2x}{4} + C\end{aligned}$$ NOTE (How to remember which sign): The power-reduction formula matches its own name: $\cos^2 x$ takes the plus sign, $\sin^2 x$ takes the minus. If you blank on it, rederive in seconds: $\cos^2 x$ comes from solving $\cos 2x = 2\cos^2 x - 1$, while $\sin^2 x$ comes from solving $\cos 2x = 1 - 2\sin^2 x$ (or from $\sin^2 x = 1 - \cos^2 x$ once you have the cosine form), rather than memorizing both. ### Sum formulas, the source of everything above The sum and difference formulas are the parents of the double-angle formulas: set $A=B$ and $\sin(A+B)$ becomes $\sin 2A$ while $\cos(A+B)$ becomes $\cos 2A$. On the AP exam you rarely apply them to an integral directly, but they let you rebuild any double-angle formula you forget, and one of them is the engine behind the derivative of sine from the definition. | Formula | Watch the sign | Calculus role | | --- | --- | --- | | $\sin(A+B) = \sin A\cos B + \cos A\sin B$ | Sine keeps the sign: a plus stays a plus. | Expanding $\sin(x+h)$ with this is the first step in deriving $(\sin x)' = \cos x$ from the limit definition. | | $\sin(A-B) = \sin A\cos B - \cos A\sin B$ | A difference turns the plus into a minus. | Set $A=B$ as a check: it correctly gives $0$. | | $\cos(A+B) = \cos A\cos B - \sin A\sin B$ | Cosine flips the sign: an outside plus becomes an inside minus. | Set $A=B$ to recover $\cos 2A=\cos^2 A-\sin^2 A$, the base double-angle form. | | $\cos(A-B) = \cos A\cos B + \sin A\sin B$ | A difference turns the minus into a plus. | The least-used of the four on the AP exam; know it for the sign pattern. | NOTE (What you can skip): The product-to-sum formulas, like $\sin A\cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)]$, belong to a full trig course and to integrals such as $\int \sin 5x\cos 3x\,dx$ that the AP exam does not ask. Leave them out. The four sum formulas here, plus the double-angle and Pythagorean rows above, are the complete working set for both AB and BC. --- # Teaching calculators ## Riemann Sum Visualizer: Rectangles, Error, and Rules URL: https://www.calclearn.org/calculators/riemann-sum-visualizer Last updated: 2026-07-22 Answer first: A Riemann sum estimates a definite integral by cutting [a,b] into n strips, multiplying each strip's height by its width, and adding. The visualizer draws rectangles and compares your total to the exact value. Left and right sums err by monotonicity; midpoint and trapezoidal sums, by concavity; all sharpen as n grows. This tool turns a definite integral into a picture you can push around. Enter a function $f$, the endpoints $a$ and $b$, and a number of subintervals $n$, then choose left, right, midpoint, or trapezoidal. The visualizer draws every strip, adds their areas into a running total, and reports the gap between that total and the exact value. Approximating area this way is Topics 6.2 and 6.3 in the CED, and it is where the definite integral is born: the area between a rate-of-change graph and the $x$-axis is the accumulated change (CHA-4.A.1). Each rectangle is a product, not just a shape. Its width is one slice of the input, $\Delta x = \frac{b-a}{n}$ for equal strips, and its height is a single function value $f(x_i)$. The product $f(x_i) \cdot \Delta x$ is that strip's area, and when $f$ is a rate it is the change accumulated over that slice. Adding the strips adds those pieces of change, which is all the summation notation is saying: $$\Delta x = \frac{b-a}{n}, \qquad S_n = \sum_{i=1}^{n} f(x_i) \cdot \Delta x$$ The four buttons differ in one decision only: where each strip reads its height. A left sum uses the left edge of every subinterval, a right sum the right edge, and a midpoint sum the center. The trapezoidal option is different in kind: it replaces each rectangle with a trapezoid whose slanted top joins the left and right function values, which is the same as averaging the left and right sums. Keep $n$ fixed and switch methods to see which strips overshoot the curve and which fall short. Drag $n$ upward and the running total settles toward one number. Narrower strips leave a thinner sliver between each flat or slanted top and the curving graph, so the total missed or extra area shrinks. That limiting number is the definite integral: as the widths of the subintervals approach $0$, the Riemann sums approach $\int_a^b f(x)\,dx$ (LIM-5.C.1, Topic 6.3). The error panel in the tool is that closing gap shown as a number, and it is why no finite $n$ ever lands exactly on the answer for a curved graph. You can usually name which side of the true value an estimate falls on before you ever compute the exact integral. Just two properties of $f$ decide it. Monotonicity settles the left and right sums; concavity settles the midpoint and trapezoidal sums. | Method | What decides it | Underestimates when | Overestimates when | | --- | --- | --- | --- | | Left sum | Monotonicity | $f$ is increasing | $f$ is decreasing | | Right sum | Monotonicity | $f$ is decreasing | $f$ is increasing | | Midpoint sum | Concavity | $f$ is concave up | $f$ is concave down | | Trapezoidal sum | Concavity | $f$ is concave down | $f$ is concave up | NOTE (The midpoint-trapezoid sandwich): For a given concavity, the midpoint and trapezoidal errors point opposite ways, so the exact integral is trapped between them. On a concave-up graph the midpoint sum sits below the true value and the trapezoidal sum sits above, so the answer lives in between. Put both methods in the visualizer at the same $n$ and watch the exact value land inside the two estimates: that is a free accuracy check you can use whenever $f$ keeps the same concavity across $[a,b]$ (no inflection point on the interval), even when you cannot find the integral by hand. ### Worked example #### Midpoint sum for a concave-up function Problem: Approximate $\int_1^3 \frac{1}{x}\,dx$ with a midpoint Riemann sum using $n = 4$ equal subintervals, then use concavity to decide whether the estimate is too high or too low and check it against the exact value. 1. Find the strip width and the grid. On $[1,3]$ with $n = 4$, $\Delta x = \frac{3-1}{4} = \frac{1}{2}$. The subintervals are $[1,1.5]$, $[1.5,2]$, $[2,2.5]$, and $[2.5,3]$. 2. A midpoint sum reads each height at the center of its subinterval, so list the four midpoints: $1.25$, $1.75$, $2.25$, and $2.75$. 3. Evaluate $f(x) = \frac{1}{x}$ at each midpoint: $f(1.25) = 0.8$, $f(1.75) \approx 0.5714$, $f(2.25) \approx 0.4444$, and $f(2.75) \approx 0.3636$ (rounded to four places). 4. Add the heights and multiply by the common width: $M_4 = \Delta x \cdot (0.8 + 0.5714 + 0.4444 + 0.3636) = 0.5 \cdot 2.1794 = 1.0897$. 5. Classify with concavity. Since $f(x) = x^{-1}$, we get $f''(x) = 2x^{-3} > 0$ on $[1,3]$, so the graph is concave up. The midpoint rectangle has the same area as the trapezoid under the tangent line at the midpoint; for a concave-up graph the curve lies above that tangent line, so the true area exceeds the rectangle and the midpoint sum underestimates. 6. Check against the exact value. $\int_1^3 \frac{1}{x}\,dx = \left[\ln x\right]_1^3 = \ln 3 - \ln 1 = \ln 3 \approx 1.0986$. The estimate $1.0897$ sits just below it, an error of about $0.0089$, and shrinking $\Delta x$ in the visualizer drives that gap toward $0$. Answer: $M_4 \approx 1.0897$, an underestimate of the exact value $\ln 3 \approx 1.0986$ because $f(x) = \frac{1}{x}$ is concave up on $[1,3]$; the error is about $0.0089$. ### FAQ Q: Do all four Riemann sums give the same answer? A: No. At any finite $n$ the left, right, midpoint, and trapezoidal sums usually give four different numbers, because each reads its strip heights from a different place. They agree only in the limit as $n \to \infty$, where all four converge to the exact integral $\int_a^b f(x)\,dx$. The exception is a straight-line $f$, where the midpoint and trapezoidal sums are already exact at every $n$ while the left and right sums still miss. Q: Why does the estimate get closer to the exact value as I increase n? A: More subintervals means each strip is narrower, so its flat or slanted top has less room to drift from the curve. The total missed-or-extra area is the sum of those small gaps, and it shrinks as the strips shrink. In the limit as the subinterval widths approach $0$, the Riemann sum equals the definite integral $\int_a^b f(x)\,dx$, which is the definition of the integral (LIM-5.C.1). Q: How does the calculator get the exact value it compares against? A: It evaluates the integral independently of your Riemann sum. For standard functions it uses an antiderivative and the Fundamental Theorem, $\int_a^b f(x)\,dx = F(b) - F(a)$, and for functions with no elementary antiderivative it falls back to a high-accuracy numerical routine that uses far more subintervals than you would by hand. The error it reports is the distance between that reference value and your estimate. Q: Are Riemann sums on the AP Calculus exam? A: Yes, in both AB and BC. They live in Topics 6.2 and 6.3, and they show up as multiple-choice questions and, very often, as a free-response part where a table of measured values asks for a left, right, midpoint, or trapezoidal approximation of an integral. Reading values off a graph or table and building the sum by hand is the tested skill, so practice the setup, not just the answer. ## Chain Rule Step Trainer: You Pick the Decomposition URL: https://www.calclearn.org/calculators/chain-rule-trainer Last updated: 2026-07-22 Answer first: The chain rule's hard part is decomposition: naming the outer and inner function before you differentiate. Use the last-operation test: whatever you compute last is the outer function, everything inside it is the inner. Differentiate the outer with the inside untouched, then multiply by the inner's derivative. The chain rule itself is one line. The step that actually decides whether you get the derivative right is decomposition: splitting a composite function like $f(x) = g(h(x))$ into an outer function $g$ and an inner function $h$ before you differentiate anything. This trainer makes you commit to that split from a set of choices, then assembles the derivative one step at a time so you can see how your pick drives every factor. That is the exact skill AP Calculus Topic 3.1 is built around: skill 1.C, choosing a procedure based on how an expression is put together rather than memorizing a formula. $$f'(x) = g'(h(x)) \cdot h'(x)$$ Read the formula as a recipe with three slots: the derivative of the outer function evaluated at the untouched inside, then multiplied by the derivative of the inside. Every slot depends on naming $g$ and $h$ correctly first, which is why the trainer asks you to choose the outer and inner function before it will build anything. To find them, use the last-operation test: imagine plugging in a number and running the arithmetic. Whatever you would do last is the outer function; everything you did before that is the inner function. NOTE (The last-operation test): For $\sqrt{7x - 1}$ you would compute $7x - 1$ first and take the square root last, so the outer function is the square root and the inner is $7x - 1$. For $e^{x^2}$ you would compute $x^2$ first and exponentiate last, so the outer function is $e^{u}$ and the inner is $x^2$. Say the split out loud before you differentiate. Once the split is named, the single habit that keeps chain-rule work clean is leaving the inside untouched while you differentiate the outside. The derivative of $(3x^2 + 5x)^4$ keeps $3x^2 + 5x$ intact as $4(3x^2 + 5x)^3$; only afterward do you multiply by the inner derivative $6x + 5$. The trainer enforces this order, so a wrong pick shows up immediately as a factor that does not belong, which is the fastest way to feel the difference between the two candidate functions. | Expression | Outer function (last step) | Inner function | | --- | --- | --- | | $(3x^2 + 5x)^4$ | raise to the 4th power, $u^4$ | $3x^2 + 5x$ | | $\sin(4x)$ | sine, $\sin u$ | $4x$ | | $e^{x^2}$ | exponential, $e^{u}$ | $x^2$ | | $\sqrt{7x - 1}$ | square root, $u^{1/2}$ | $7x - 1$ | | $\sin^2 x$ | square, $u^2$ | $\sin x$ | The last row is the classic trap: $\sin^2 x$ means $(\sin x)^2$, so squaring is the outer function and $\sin x$ is the inner, which is the opposite decomposition from $\sin(x^2)$. Working several of these in the trainer trains your eye to read the notation before you commit. Aim to name the outer and inner function correctly on sight; the assembly that follows is mechanical once the split is right, and getting it solid here pays off in implicit differentiation (Topic 3.2), related rates (Topic 4.4), and $u$-substitution, all of which are the chain rule wearing a different hat. ### Worked example #### Decompose, then assemble a square-root composition Problem: Differentiate $f(x) = \sqrt{x^2 + 1}$. 1. Rewrite the root as a power so the outer function is easy to name: $\sqrt{x^2 + 1} = (x^2 + 1)^{1/2}$. This is the same rewrite the trainer prompts before you choose, because a radical hides the fact that the last operation is a power. 2. Run the last-operation test. Evaluating at a number, you would compute $x^2 + 1$ first and raise it to the $1/2$ power last. So the outer function is $g(u) = u^{1/2}$ and the inner function is $h(x) = x^2 + 1$, giving $f(x) = g(h(x))$. 3. Differentiate the outer function, leaving the inside untouched. Since $g(u) = u^{1/2}$, we have $g'(u) = \tfrac{1}{2} u^{-1/2}$, so $g'(h(x)) = \tfrac{1}{2}(x^2 + 1)^{-1/2}$. Notice the inside stays exactly $x^2 + 1$. 4. Differentiate the inner function on its own: $h'(x) = 2x$. 5. Multiply the two pieces per the chain rule: $f'(x) = g'(h(x)) \cdot h'(x) = \tfrac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x$. 6. Simplify. The $\tfrac{1}{2}$ and the $2x$ combine to $x$, and the negative exponent moves the factor into a denominator: $f'(x) = \dfrac{x}{\sqrt{x^2 + 1}}$. Answer: $f'(x) = \dfrac{x}{\sqrt{x^2 + 1}}$ ### FAQ Q: How do I decide which function is the outer and which is the inner? A: Use the last-operation test. Imagine evaluating the expression at a number and watch the order of operations: whatever you would do last is the outer function, and everything you computed before that is the inner function. For $\cos(3x - 1)$ you subtract inside first and take cosine last, so cosine is outer and $3x - 1$ is inner. Q: What happens if I pick the outer and inner functions backwards? A: You differentiate the wrong function and the factors come out mismatched. If you treated $\sin(x^2)$ as though squaring were the outer step (as in $\sin^2 x$), you would differentiate $(\sin x)^2$ and get $2\sin x \cos x$, instead of the correct derivative $2x\cos(x^2)$, which is a different function entirely. The trainer surfaces this right away so you can see why the split has to come first. Q: Do I need to rewrite roots and reciprocals before decomposing? A: Yes, and it makes the outer function obvious. Rewrite $\sqrt{u}$ as $u^{1/2}$ and $\dfrac{1}{u}$ as $u^{-1}$ so the last operation reads as a power. Then $g(u) = u^{1/2}$ or $g(u) = u^{-1}$ is the outer function and the power rule handles $g'$, with the chain rule multiplying by the inner derivative. ## U-Substitution Helper: Pick u, See Why It Works URL: https://www.calclearn.org/calculators/u-substitution-helper Last updated: 2026-07-22 Answer first: Pick u to be the inside function whose derivative already appears in the integrand as a factor. Test a candidate by differentiating it: if du (up to a constant) is present, the choice works and the integral collapses; otherwise pick a different u. This tool runs that test and hints instead of solving. This helper drills the single decision that makes or breaks a substitution: which part of the integrand to call $u$. You pick a candidate from the pieces on screen, and the tool differentiates it, checks whether $du$ is available in the integrand, and rewrites the integral in $u$ when the choice is sound. When the choice is poor it stops and returns a hint instead of an answer, so the goal is to learn the reasoning, not to be handed a solved integral. Substitution is CED Topic 6.9 (FUN-6.D.1), and it shows up on both the AB and BC exams, usually as one step buried inside a larger accumulation or area problem. $$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du, \qquad u = g(x)$$ Substitution is the chain rule run backward. The chain rule turns $F(g(x))$ into $F'(g(x)) \cdot g'(x)$, so an integral that carries a composite piece next to the derivative of its inner function is built to be un-chained. Every $u$ you would ever pick comes from the same short list of spots, all of them the inside of a composite: the base of a power like $(x^2+3)^4$, the expression under a root, the exponent on $e$, the argument of a $\sin$, $\cos$, or $\tan$, or a denominator. Each inner function is a candidate. The tool surfaces those candidates; your job is to decide which one has its derivative sitting elsewhere in the integrand. NOTE (The test the tool runs): Differentiate your candidate. If the result, times any constant, is already a factor in the integrand, then $du$ can be formed and the substitution works. If differentiating produces a factor that appears nowhere, that candidate fails and you get a hint. When your pick passes, read what the rewrite shows: every $x$ disappears, including inside the $dx$, and the integral collapses into a basic form in $u$. A leftover constant is harmless, because you fix it by multiplying and dividing. If $du = 2x\,dx$ and the integral only has $x\,dx$, then $x\,dx = \frac{1}{2}\,du$. A leftover variable is not harmless: if an $x$ still survives after substituting, the rewrite is incomplete and the choice usually has to change. Two failure signals trigger a hint. The derivative is missing: set $u = x^2$ in $\int e^{x^2}\,dx$ and $du = 2x\,dx$ needs an $x$ that is not there, so the substitution stalls (this integrand has no elementary antiderivative at all). A stray variable is left behind: the rewrite still carries an $x$ that $du$ cannot absorb. Reading which signal fired tells you what to do next, either try a different candidate or, in the stray-variable case, solve $u = g(x)$ for $x$ and substitute that too. ### Worked example #### Testing candidates: why u equals x^4 beats u equals x^3 Problem: Evaluate $\int x^3 \cos(x^4)\,dx$. 1. List the candidates the way the tool does. The integrand has two pieces, the factor $x^3$ and the composite $\cos(x^4)$. The plausible candidates for $u$ are $u = x^3$ (the standalone factor) and $u = x^4$ (the inner function of the composite). Test each by differentiating and checking whether its $du$ is present. 2. Test $u = x^3$. Then $\frac{du}{dx} = 3x^2$, so $du = 3x^2\,dx$. The factor $x^2$ is available, since $x^3 = x \cdot x^2$, but pulling out $x^2\,dx$ leaves a stray $x$: $x^3\,dx = \frac{x}{3}\,du$. That leftover $x$ cannot be written in terms of $u = x^3$ (the composite's argument $x^4$ is not a function of $x^3$ either), so the rewrite still carries an $x$. This is the stray-variable signal, and the tool returns a hint rather than a clean rewrite. 3. Test $u = x^4$. Then $\frac{du}{dx} = 4x^3$, so $du = 4x^3\,dx$. The factor $x^3\,dx$ is exactly what sits next to the cosine, matched up to the constant $4$. Solve for that piece: $x^3\,dx = \frac{1}{4}\,du$. This candidate passes. 4. Rewrite the integral entirely in $u$: $\int x^3 \cos(x^4)\,dx = \int \cos(x^4)\cdot x^3\,dx = \int \cos u \cdot \frac{1}{4}\,du$. No $x$ remains anywhere, which confirms the substitution is complete. 5. Factor the constant out front: $\int \cos u \cdot \frac{1}{4}\,du = \frac{1}{4}\int \cos u\,du$. 6. Integrate in $u$: $\frac{1}{4}\int \cos u\,du = \frac{1}{4}\sin u$. 7. Back-substitute $u = x^4$ and add the constant: $\frac{1}{4}\sin(x^4) + C$. 8. Check by differentiating: $\frac{d}{dx}\left[\frac{1}{4}\sin(x^4)\right] = \frac{1}{4}\cos(x^4)\cdot 4x^3 = x^3\cos(x^4)$, which is the original integrand. Answer: $\dfrac{1}{4}\sin(x^4) + C$ ### FAQ Q: How do you choose u in u-substitution? A: Pick the inner function of a composite whose derivative also appears in the integrand. Test a candidate by differentiating it: if the result, times any constant, is already a factor in the integral, $du$ can be formed and the choice works. The usual candidates are bases of powers, radicands, exponents of $e$, trig arguments, and denominators. Q: Why did the helper flag my u with a hint instead of solving? A: Because your candidate did not pass the test. Either its derivative is missing (differentiating produced a factor that is nowhere in the integrand, so $du$ cannot be built), or the rewrite left a stray $x$ that $du$ cannot absorb. The hint names which signal fired so you know whether to try another candidate or to solve $u = g(x)$ for $x$. Q: Does it matter which candidate I try first? A: No. The differentiate-and-check test is quick, and only the candidate whose derivative is present produces a clean rewrite with no $x$ left. Try any candidate; the tool will pass the good one and hint on the rest. In the stray-variable case you may also need to solve $u = g(x)$ for $x$ before every $x$ clears. ## Series Convergence Test Chooser: Which Test and Why (BC) URL: https://www.calclearn.org/calculators/convergence-test-chooser Last updated: 2026-07-22 Answer first: Which convergence test to use is set by the shape of the terms. Start with the nth-term test: if the terms do not approach 0, the series diverges. If they do, match the shape: geometric and p-series by sight, alternating signs to the alternating series test, factorials to the ratio test, rational terms to comparison. This tool walks the convergence decision tree one question at a time. You pick or enter a series and it asks the same questions a strong BC student asks, in order, then explains why each test either applies to your series or fails on it. That reasoning is the whole skill of Unit 10 (Topics 10.1 through 10.9): a solver can tell you a series converges, but the exam rewards choosing the right test and justifying it. Only six named tests are assessed on the BC exam, and the tool uses exactly those, plus the geometric and p-series facts you recognize on sight. There is no single convergence test. Instead you run a short list of them in a fixed order, and the shape of the general term $a_n$ decides which one settles the question. The order matters because the cheapest check comes first. Every series starts on the same rung, the nth-term test, and you only move deeper into the tree when a test comes back inconclusive. NOTE (The nth-term test only proves divergence): If $\lim_{n \to \infty} a_n \ne 0$, the series diverges, full stop (Topic 10.3). But $\lim_{n \to \infty} a_n = 0$ proves nothing on its own: the harmonic series $\sum \frac{1}{n}$ has terms going to $0$ and still diverges. A zero limit just clears the series to move on to a real test, so never write 'the terms go to $0$, therefore it converges.' Once the nth-term test is inconclusive, the shape of $a_n$ points to the test that will finish the job. Read the left column as what your term looks like, and the tool selects the row that matches. | When the term looks like | The tool reaches for | Because | | --- | --- | --- | | $\lim_{n \to \infty} a_n \ne 0$ | nth-term test (Topic 10.3) | The series diverges at once; no further test is needed. | | $a r^n$, a constant ratio between terms | Geometric series (Topic 10.2) | It converges to $\frac{a}{1-r}$ exactly when $|r| < 1$. | | $\frac{1}{n^p}$, a single power of $n$ | p-series (Topic 10.5) | It converges exactly when $p > 1$; the harmonic case $p = 1$ diverges. | | Signs alternate, $(-1)^n b_n$ | Alternating series test (Topic 10.7) | It converges if the positive terms $b_n$ decrease monotonically to $0$. | | Factorials or $n$-th powers | Ratio test (Topic 10.8) | The ratio $\frac{a_{n+1}}{a_n}$ makes those pieces cancel. | | A rational or algebraic function of $n$ | Comparison or limit comparison (Topic 10.6) | Match it to the p-series its leading powers resemble. | | A positive, continuous, decreasing term you can integrate | Integral test (Topic 10.4) | The series and $\int_1^\infty f(x)\,dx$ converge together. | Three of those rows deserve a sharper rule. The ratio test is the specialist for factorials and $n$-th powers, because $\frac{a_{n+1}}{a_n}$ makes those pieces cancel; on a plain rational function of $n$ its limit is $1$ and tells you nothing, so do not reach for it there. Comparison and limit comparison are the tools for rational and algebraic terms: match the term to the p-series its leading powers resemble, then compare. The integral test is the fallback when a term is positive, continuous, decreasing, and something you can actually antidifferentiate, like $\frac{1}{n \ln n}$. Two shapes never need a test at all, because their rules are exact: $$\sum_{n=0}^{\infty} a r^n \text{ converges} \iff |r| < 1 \qquad \sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges} \iff p > 1$$ Once a series passes the alternating series test, one more question follows: is it absolutely or conditionally convergent (Topic 10.9)? Take the series of absolute values and run a positive-term test on it. If that converges, the original converges absolutely, and absolute convergence guarantees convergence. If the absolute series diverges but the alternating series test still passes, the convergence is only conditional, like the alternating harmonic series $\sum \frac{(-1)^{n+1}}{n}$. ### Worked example #### Choosing a test for a rational-term series Problem: Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{3n^2 + 2}{n^4 + n + 1}$ converges or diverges, and name the test that decides it. 1. Start on the bottom rung, the nth-term test. The denominator has degree $4$ and the numerator degree $2$, so $\lim_{n \to \infty} \frac{3n^2 + 2}{n^4 + n + 1} = 0$. A zero limit never proves convergence, so this test is inconclusive and you keep going (Topic 10.3). 2. Rule out the by-sight forms. Every term is positive, so it is not alternating (Topic 10.7 is off the table). There is no constant ratio between terms, so it is not geometric (Topic 10.2), and it is not a single power $\frac{1}{n^p}$, so it is not a bare p-series (Topic 10.5). No factorials or $n$-th powers appear, and on a rational function of $n$ the ratio test's limit is $1$, which is inconclusive, so skip it (Topic 10.8). 3. What is left is a rational function of $n$, which points to comparison (Topic 10.6). Pick the partner from the leading powers: for large $n$ the term behaves like $\frac{3n^2}{n^4} = \frac{3}{n^2}$, so compare against $b_n = \frac{1}{n^2}$, a p-series with $p = 2$. Because a clean term-by-term inequality is awkward here, use the limit comparison test, which only needs the limit of the ratio. 4. Set up the limit comparison ratio, a fraction divided by a fraction: $\frac{a_n}{b_n} = \frac{\frac{3n^2 + 2}{n^4 + n + 1}}{\frac{1}{n^2}}$. 5. Multiply numerator and denominator through by $n^2$ to clear the complex fraction: $\frac{n^2(3n^2 + 2)}{n^4 + n + 1} = \frac{3n^4 + 2n^2}{n^4 + n + 1}$. 6. Evaluate the limit by dividing top and bottom by $n^4$: $\lim_{n \to \infty} \frac{3 + \frac{2}{n^2}}{1 + \frac{1}{n^3} + \frac{1}{n^4}}$. 7. Every term with $n$ in a denominator vanishes, leaving $\frac{3}{1} = 3$, a finite, positive number. 8. Conclude. The comparison series $\sum \frac{1}{n^2}$ is a convergent p-series ($p = 2 > 1$). When the limit comparison ratio is finite and positive, both series share the same fate, so the given series converges (Topic 10.6). Answer: The series converges, by limit comparison with the convergent p-series sum of 1/n^2 (the ratio has finite positive limit 3). ### FAQ Q: What is the difference between the comparison test and the limit comparison test? A: The direct comparison test needs an inequality: if $0 \le a_n \le b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges, and if $a_n \ge b_n \ge 0$ and $\sum b_n$ diverges, then $\sum a_n$ diverges. The limit comparison test drops the inequality and only asks for $\lim_{n \to \infty} \frac{a_n}{b_n}$: if that limit is a finite positive number, both series share the same fate. Use limit comparison when the term-by-term inequality is awkward to prove, which is most rational-function cases (Topic 10.6). Q: Does the nth-term test prove a series converges? A: No. It is a one-directional test. If $\lim_{n \to \infty} a_n \ne 0$ the series diverges, but $\lim_{n \to \infty} a_n = 0$ tells you nothing by itself. The harmonic series $\sum \frac{1}{n}$ has terms going to $0$ yet diverges, so a zero limit only earns the series a real convergence test (Topic 10.3). Q: When should I use the ratio test? A: Reach for the ratio test when the term contains factorials or $n$-th powers such as $r^n$ or $n^n$, because $\frac{a_{n+1}}{a_n}$ cancels those pieces cleanly. If the limit $L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$ the series converges absolutely, if $L > 1$ it diverges, and if $L = 1$ the test is inconclusive and you switch to another test. On a plain rational function of $n$ the ratio limit is always $1$, so use comparison there instead (Topic 10.8). Q: Which convergence tests are on the AP Calculus BC exam? A: The CED assesses exactly six: the nth-term test for divergence, the integral test, the comparison test, the limit comparison test, the alternating series test, and the ratio test (the Topic 10.8 exclusion statement). You also need the geometric series and p-series convergence rules from Topics 10.2 and 10.5. The root test is not assessed, so you will never need it on the exam. ## Limit Solver (Teaching Mode): Which Method and Why URL: https://www.calclearn.org/calculators/limit-method-chooser Last updated: 2026-07-22 Answer first: Which method a limit needs is decided by the form direct substitution returns. A real number is the answer; a nonzero number over 0 is an infinite limit or does not exist; 0/0 is indeterminate, so climb the ladder: factor polynomials, use the conjugate for radicals, match sin x / x trig shapes, and use L'Hopital last. This tool does more than hand you the value of a limit. For each curated limit it shows which method the limit needs and why, which is the exact skill Topic 1.7 tests: selecting a procedure. Every limit starts the same way, with direct substitution, and the form substitution returns is what picks the method. Think of the methods as rungs on a ladder. You always step on the bottom rung first (substitute), and you only climb higher when the form you get back is indeterminate. The table below is the whole decision: read what substitution returns on the left, and the method it selects is on the right. | After direct substitution you get | What it means | Method the tool selects | | --- | --- | --- | | A real number | The function is continuous there (Topic 1.12) | Stop. That number is the limit. | | $\frac{0}{0}$ with polynomials | A common factor is hiding | Factor and cancel (Topic 1.6) | | $\frac{0}{0}$ with a radical | A root is blocking the cancellation | Multiply by the conjugate (Topic 1.6) | | $\frac{0}{0}$ from a $\frac{\sin x}{x}$ shape | A special trig limit | Rewrite to a known trig limit (Topic 1.8) | | $\frac{0}{0}$ or $\frac{\infty}{\infty}$ that resists algebra | Still indeterminate | L'Hopital's Rule (Topic 4.7) | | A nonzero number over $0$, like $\frac{5}{0}$ | Vertical asymptote | Infinite limit or does not exist (Topic 1.14) | NOTE: $\frac{0}{0}$ is an indeterminate form, which is a signal to keep working, not an answer. It never means the limit equals $0$, and it does not by itself mean the limit does not exist. It only tells you a common factor or a special trig shape is hiding. The sharpest selection is factor versus conjugate, and the expression tells you which. If both the top and bottom are polynomials, factor each one, cancel the shared $(x - c)$, and substitute into what remains. If a square root is trapped inside the $\frac{0}{0}$, factoring cannot clear it, so multiply by the conjugate of the radical to turn a difference like $\sqrt{x+4} - 2$ into a difference of squares. Quick rule: polynomials mean factor, a stubborn root means conjugate. Some $\frac{0}{0}$ limits will not factor and have no radical, because the obstacle is trigonometric. When you see a sine divided by its own argument, or something you can reshape into that, match it to one of the two limits the squeeze theorem gives you (Topic 1.8): $$\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0$$ L'Hopital's Rule is the top rung, and you only reach it when algebra stalls. It applies to exactly two forms, $\frac{0}{0}$ and $\frac{\infty}{\infty}$, and it uses derivatives, so it lands later in the course at Topic 4.7 (both AB and BC). Verify the indeterminate form first: if substitution already gives a form that is not indeterminate, such as $\frac{5}{0}$, the rule does not apply. Other forms such as $\infty - \infty$ are not assessed on the AP exam. ### Worked example #### Choosing the method for a 0/0 limit Problem: Evaluate $\lim_{x \to 7} \dfrac{\sqrt{x+2} - 3}{x - 7}$. 1. Start on the bottom rung: direct substitution. Substituting $x = 7$ gives $\dfrac{\sqrt{9} - 3}{7 - 7} = \dfrac{3 - 3}{0} = \dfrac{0}{0}$. That is indeterminate, so substitution failed and you keep climbing (Topic 1.12). 2. Select the rung by reading the form. You have $\dfrac{0}{0}$ with a square root sitting in the numerator. A radical blocks ordinary factoring, so the ladder sends you to the conjugate, not to factoring (Topic 1.6). 3. Multiply the numerator and denominator by the conjugate of the radical, $\sqrt{x+2} + 3$: $\dfrac{\sqrt{x+2} - 3}{x - 7} \cdot \dfrac{\sqrt{x+2} + 3}{\sqrt{x+2} + 3}$. 4. The numerator is a difference of squares: $(\sqrt{x+2})^2 - 3^2 = (x + 2) - 9 = x - 7$. The expression becomes $\dfrac{x - 7}{(x - 7)(\sqrt{x+2} + 3)}$. 5. Cancel the common factor $x - 7$. This is legal because a limit only looks at $x$ near $7$, never at $7$ itself, so $x - 7 \ne 0$ for every point that matters. You are left with $\dfrac{1}{\sqrt{x+2} + 3}$. 6. Substitution works now, so drop back to the bottom rung on the simplified form: $\dfrac{1}{\sqrt{9} + 3} = \dfrac{1}{3 + 3} = \dfrac{1}{6}$. Answer: $\dfrac{1}{6}$ ### FAQ Q: Do I always have to try direct substitution first? A: Yes. Substitution is the bottom rung and often the entire problem: if the function is continuous at the point, the value you get is the limit (Topic 1.12). You only climb to factoring, conjugates, trig limits, or L'Hopital when substitution returns an indeterminate $\frac{0}{0}$. Q: How do I know whether to factor or use the conjugate? A: Look for a square root. If the $\frac{0}{0}$ expression is built from polynomials, factor and cancel. If a radical is trapped inside, factoring cannot remove it, so multiply by the conjugate to create a difference of squares, then cancel (Topic 1.6). Q: Can I just use L'Hopital's Rule on every 0/0 limit? A: It works on $\frac{0}{0}$ and $\frac{\infty}{\infty}$, but it uses derivatives and appears later at Topic 4.7, so early limit questions expect the algebraic methods instead. Always verify the form first: if substitution gives a non-indeterminate form like $\frac{5}{0}$, the rule does not apply and using it gives a wrong answer. Q: What if substitution gives a number over zero, like 5/0? A: That is not indeterminate. A nonzero number over zero signals a vertical asymptote, so the limit is $+\infty$, $-\infty$, or does not exist depending on the sign from each side (Topic 1.14). Only $\frac{0}{0}$ calls for the algebra rungs. ## Definite Integral Step Tool: FTC, Shown Honestly URL: https://www.calclearn.org/calculators/ftc-step-tool Last updated: 2026-07-22 Answer first: To evaluate a definite integral, find any antiderivative F of the integrand, then compute F(top) minus F(bottom). That is the Fundamental Theorem of Calculus: the integral from a to b of f equals F(b) - F(a). The hard part is finding the antiderivative; the evaluation is just substitute, substitute, subtract. This tool takes a curated set of definite integrals and shows the evaluation the way you have to reproduce it on the no-calculator sections: find an antiderivative, substitute the top bound, substitute the bottom bound, and subtract. A solver hands you a number and hides all three steps. Here every line stays on the table, so you learn the structure instead of just the answer. $$\int_a^b f(x)\,dx = F(b) - F(a), \quad \text{where } F'(x) = f(x)$$ The Fundamental Theorem of Calculus (Topic 6.7, essential knowledge FUN-6.B.3) applies whenever $f$ is continuous on the closed interval $[a, b]$ and you can find an antiderivative $F$. Recognizing that the FTC is the tool for the job is the easy part: numbers on the integral sign mean a definite integral, and the FTC is how you turn that into an exact value. The real work is finding $F$, which is differentiation run backward. NOTE (Why the constant of integration disappears): An indefinite integral needs $+\,C$, but a definite integral does not. If you carried the constant, you would compute $(F(b) + C) - (F(a) + C)$, and the $C$ cancels. So pick the simplest antiderivative, the one with $C = 0$, and evaluate that. 1. Confirm a definite integral: the sign carries a lower bound $a$ and an upper bound $b$, and the integrand is continuous on $[a, b]$. 2. Find any one antiderivative $F$ of the integrand by reversing your differentiation rules. 3. Substitute the upper bound to get $F(b)$. 4. Substitute the lower bound to get $F(a)$. 5. Subtract in that order, $F(b) - F(a)$: upper minus lower, never the reverse. NOTE (The mistake that costs the most points): Reversing the subtraction. $F(a) - F(b)$ gives the negative of the correct answer, and on a free-response question a sign flip can cascade into every later part. The bracket notation $\left[ F(x) \right]_a^b$ exists to remind you: top value first, then subtract the bottom value. When the antiderivative is not obvious, the FTC still runs these same steps; you just need a technique to find $F$ first. If the integrand is a composition with the inner function's derivative present, try u-substitution (Topic 6.9). For a product of unlike functions, BC students reach for integration by parts (Topic 6.11, BC only). Choosing that technique is a separate decision from the evaluation you drill here, which is why this tool keeps the two apart: the antidifferentiation is kept routine (reverse power rule, or the antiderivative is supplied), so you can focus on the bracket-and-subtract mechanics that every FTC problem shares. ### Worked example #### Evaluating a definite integral with the FTC Problem: Evaluate $\int_1^3 (3x^2 + 2)\,dx$. 1. Check that the theorem applies. The integrand $3x^2 + 2$ is a polynomial, so it is continuous on $[1, 3]$, and both bounds are finite. This is a definite integral of a continuous function, so the Fundamental Theorem of Calculus gives an exact value: find an antiderivative and evaluate it at the bounds. 2. Find an antiderivative using the reverse power rule, term by term. An antiderivative of $3x^2$ is $x^3$, and an antiderivative of the constant $2$ is $2x$, so take $F(x) = x^3 + 2x$. Verify it: $F'(x) = 3x^2 + 2$, which matches the integrand, so $F$ is correct. No $+\,C$ is needed, because it would cancel in the subtraction. 3. Set up the evaluation bracket. Write the antiderivative in brackets with the bounds attached: $\left[ x^3 + 2x \right]_1^3$, which is shorthand for $F(3) - F(1)$. 4. Substitute the upper bound $x = 3$: $F(3) = 3^3 + 2(3) = 27 + 6 = 33$. 5. Substitute the lower bound $x = 1$: $F(1) = 1^3 + 2(1) = 1 + 2 = 3$. 6. Subtract, upper minus lower: $F(3) - F(1) = 33 - 3 = 30$. Answer: $\int_1^3 (3x^2 + 2)\,dx = 30$. ### FAQ Q: Do you need a +C when evaluating a definite integral? A: No. The constant is only for indefinite integrals. In a definite integral you compute $(F(b) + C) - (F(a) + C)$, and the $C$ cancels, so it never affects the answer. Use the antiderivative with $C = 0$. Q: What is the difference between the two parts of the Fundamental Theorem of Calculus? A: The first part (Topic 6.4) is about accumulation functions: if $g(x) = \int_a^x f(t)\,dt$, then $g'(x) = f(x)$, so differentiating an integral gives the integrand back. The second part (Topic 6.7) is the evaluation rule this tool shows: $\int_a^b f(x)\,dx = F(b) - F(a)$. This tool teaches the second part. Q: What is the difference between a definite and an indefinite integral? A: A definite integral has bounds and evaluates to a single number, $F(b) - F(a)$. An indefinite integral has no bounds and evaluates to a whole family of antiderivatives, $F(x) + C$. The FTC is the bridge: you find the indefinite integral (an antiderivative), then evaluate it at the two bounds to get the definite value. Q: Can you always use the FTC to evaluate a definite integral? A: Only when you can find an antiderivative in closed form. Many continuous functions, such as $e^{-x^2}$, have no elementary antiderivative (FUN-6.C.3), so you cannot use the FTC to produce an exact value by hand. In those cases you approximate the definite integral with a Riemann sum or another numerical method instead. ## Tangent and Normal Line Finder (Teaching Mode) URL: https://www.calclearn.org/calculators/tangent-line-calculator Last updated: 2026-07-22 Answer first: The tangent line at x = a has slope f'(a) and passes through the point (a, f(a)), so its equation is y = f'(a)(x - a) + f(a). The normal line is perpendicular there, so its slope is the negative reciprocal -1/f'(a). Method: find f'(x), evaluate it at a to get a number, then use point-slope form. A tangent line touches a curve at a single point and matches its steepness there; a normal line passes through that same point at a right angle to the tangent. To build either one you need two things: a slope and a point. The point is $(a, f(a))$, and the slope is $f'(a)$, the derivative evaluated at $x = a$. That single fact, that the derivative at a point equals the slope of the tangent line there, is AP Topic 2.2 (CHA-2.C.1), and it is what this tool makes visible: it computes $f'(x)$, evaluates the slope at the point you pick, and assembles both the tangent and the normal equation from point-slope form. Here is why the tangent slope is the derivative and not a separate formula you have to memorize. The average rate of change of $f$ between $x = a$ and a nearby point $x$ is the difference quotient: $$\frac{f(x) - f(a)}{x - a}$$ This is exactly the slope of the secant line through those two points (Topic 2.1). As the second point slides toward $x = a$, the secant pivots into the tangent, and that limit of secant slopes is the definition of $f'(a)$. So the tangent slope and the derivative are the same number by construction. That same tangent line doubles as the local linear approximation of $f$ near $x = a$ (Topic 4.6), so the equation you build here gets reused when you estimate function values. $$y - f(a) = f'(a)(x - a) \quad\Longrightarrow\quad y = f'(a)(x - a) + f(a)$$ The normal line shares the point $(a, f(a))$ but runs perpendicular to the tangent. Perpendicular lines have slopes that are negative reciprocals, so if the tangent slope is $f'(a)$, the normal slope is $-\frac{1}{f'(a)}$. Everything else is identical: same point, point-slope form, solve for $y$. The one case to watch is a horizontal tangent, where $f'(a) = 0$: the normal is then the vertical line $x = a$, because $-\frac{1}{0}$ is not a number. $$y - f(a) = -\frac{1}{f'(a)}(x - a)$$ NOTE (Evaluate the slope to a number first): The most common lost point on this problem type is writing the slope as $f'(x)$ with $x$ still in it. The slope of one specific tangent line is a single number, so differentiate to get $f'(x)$, then substitute $x = a$ before you write the equation. On calculator-active AP questions you may evaluate $f'(a)$ numerically, but you still report the line, not the derivative. | Behavior at $x = a$ | Tangent slope $f'(a)$ | What the two lines look like | | --- | --- | --- | | $f'(a)$ is a nonzero number | that number | slanted tangent; normal slope $-\frac{1}{f'(a)}$ | | $f'(a) = 0$: horizontal tangent | $0$ | tangent is $y = f(a)$; normal is vertical $x = a$ | | $f'(a)$ does not exist: corner or cusp | undefined | no tangent line by the standard method (Topic 2.4) | | $f'(a)$ does not exist: vertical tangent | undefined | the tangent line exists but is vertical, $x = a$, with no slope, so the slope-based point-slope method cannot build it (Topic 2.4) | ### Worked example #### Tangent and normal line to a parabola Problem: Find the tangent line and the normal line to $f(x) = x^2 + 1$ at $x = 2$. 1. Find the derivative function with the power rule (Topic 2.5). For $f(x) = x^2 + 1$, the derivative is $f'(x) = 2x$. This is a function of $x$, not yet a slope. 2. Evaluate the derivative at the point to turn it into a slope. At $x = 2$, $f'(2) = 2 \cdot 2 = 4$. This $4$ is the slope of the tangent line, because the derivative at a point is the slope of the tangent there (CHA-2.C.1). Do not leave $x$ in the slope. 3. Find the point of tangency on the curve by evaluating $f$, not $f'$: $f(2) = 2^2 + 1 = 5$. So both lines pass through $(2, 5)$. 4. Write the tangent line in point-slope form with slope $4$ and point $(2, 5)$: $y - 5 = 4(x - 2)$. Distribute and solve for $y$: $y = 4x - 8 + 5 = 4x - 3$. 5. Get the normal slope as the negative reciprocal of the tangent slope: $-\frac{1}{4}$. The normal line is perpendicular to the tangent at $(2, 5)$. 6. Write the normal line in point-slope form with slope $-\frac{1}{4}$ and the same point $(2, 5)$: $y - 5 = -\frac{1}{4}(x - 2)$. Solve for $y$: $y = -\frac{1}{4}x + \frac{1}{2} + 5 = -\frac{1}{4}x + \frac{11}{2}$. Answer: Tangent line: $y = 4x - 3$. Normal line: $y = -\frac{1}{4}x + \frac{11}{2}$. Both pass through the point of tangency $(2, 5)$. ### FAQ Q: Why is the slope of the tangent line the derivative? A: Because the derivative is defined as the limit of average rates of change. The slope between $x = a$ and a nearby point is the secant slope (the difference quotient from Topic 2.1); as the nearby point approaches $a$, the secant lines approach the tangent line, and their slopes approach $f'(a)$. The tangent slope is that limit, so it equals $f'(a)$ (Topic 2.2, CHA-2.C.1). Q: How do I find the normal line once I have the tangent? A: Keep the same point $(a, f(a))$ and use the negative reciprocal of the tangent slope. If the tangent slope is $f'(a)$, the normal slope is $-\frac{1}{f'(a)}$, and the normal line is $y - f(a) = -\frac{1}{f'(a)}(x - a)$. Perpendicular slopes always multiply to $-1$, which is what negative reciprocal means. Q: What if the derivative is zero at the point? A: Then the tangent is horizontal: $f'(a) = 0$ makes the tangent line $y = f(a)$, a flat line through the point. The normal is perpendicular to horizontal, so it is the vertical line $x = a$. You cannot write it as $y = mx + b$ because $-\frac{1}{0}$ has no value, so state the normal as $x = a$. Q: Do I plug in the point before or after differentiating? A: After. Differentiate first to get the whole function $f'(x)$, then substitute $x = a$ to get the slope as one number. Substituting into $f(x)$ too early loses the variable you need to differentiate, and it is another common mistake that produces a wrong slope on this problem type. ## Taylor Series Builder: Coefficients from Derivatives URL: https://www.calclearn.org/calculators/taylor-series-builder Last updated: 2026-07-22 Answer first: A Taylor series is an infinite polynomial (a power series) reverse-engineered so its derivatives match f at a center a: the coefficient of the (x-a)^n term is the nth derivative of f at a divided by n factorial. A Maclaurin series is just this centered at 0. Grind the derivatives, or reuse a known series. This tool builds a Taylor polynomial one derivative at a time. Pick a function $f$, a center $a$, and a degree $n$; the builder fills a table of derivatives $f, f', f'', \dots$, evaluates each one at the center, divides by a factorial, and drops the results in as coefficients. Then it graphs the polynomial on top of $f$ so you can watch it hug the curve near $x = a$. Finding these approximations is CED Topic 10.11, and the one fact doing all the work is that the coefficient of the degree-$n$ term is $\frac{f^{(n)}(a)}{n!}$ (LIM-8.A.1). The reason that formula works is worth seeing, because it is the whole idea of the tool. A Taylor polynomial is reverse-engineered to agree with $f$ at the center as closely as a polynomial of its degree can: it matches the value $f(a)$, the slope $f'(a)$, the concavity $f''(a)$, and every higher derivative up to order $n$, all at $x = a$. Each term you add pins down exactly one more derivative. That is why more terms mean a tighter fit near the center, and why the fit is sharpest right at $x = a$ and loosens as you move away. $$P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n$$ Where do the factorial and the power $(x-a)^k$ come from? Differentiate a single term $c_k(x-a)^k$ a total of $k$ times: every term below it has already vanished, every term above it still carries a factor of $(x-a)$ that dies at $x = a$, and the term itself becomes $k!\,c_k$. Setting that equal to $f^{(k)}(a)$ forces $c_k = \frac{f^{(k)}(a)}{k!}$. The tool's derivative table is exactly this calculation laid out column by column: one row per derivative order, its value at the center, and the factorial you divide by to get the coefficient. A Maclaurin series is not a different object; it is a Taylor series with the center fixed at $a = 0$. Set the center to $0$ in the builder and two things simplify: every $(x-a)^k$ collapses to a clean power $x^k$, and you evaluate each derivative at $0$, usually the easiest input to plug in. That is why the three series worth memorizing, $e^x$, $\sin x$, and $\cos x$, are always quoted as Maclaurin series (LIM-8.F.2). Centering anywhere else, say $a = 1$ for $\ln x$, gives a genuine Taylor series, and that is where the from-derivatives method earns its keep. NOTE (Two ways to get a series: pick the shorter one): Grinding out derivatives is the definition, not always the fast route, and knowing when to skip it is the real skill (Topic 10.15). Before you build a table, check whether $f$ is a known Maclaurin series in disguise. If $f$ is $e^x$, $\sin x$, $\cos x$, or $\frac{1}{1-x}$ after a substitution, a constant multiple, a derivative, or an integral, start from the known series instead: to get the series for $e^{-x^2}$, substitute $-x^2$ into the series for $e^x$ rather than differentiating $e^{-x^2}$ four times. Build from derivatives when the center is not $0$, or when $f$ does not reduce to one of the four standard series. | Function | Maclaurin series | Reach for it when | | --- | --- | --- | | $e^x$ | $\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \cdots$ | any exponential, or a substitution like $e^{-x^2}$ | | $\sin x$ | $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \cdots$ | odd-powered, sign-alternating targets | | $\cos x$ | $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \cdots$ | even-powered, sign-alternating targets | | $\frac{1}{1-x}$ | $\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots$ for $|x| < 1$ | rational functions, via algebra or substitution | ### Worked example #### Degree-3 Taylor polynomial for ln x centered at a = 1 Problem: Build the third-degree Taylor polynomial $P_3(x)$ for $f(x) = \ln x$ centered at $a = 1$, taking each coefficient straight from a derivative. 1. Set up the coefficient formula at this center. With $a = 1$, the polynomial is $P_3(x) = \sum_{k=0}^{3} \frac{f^{(k)}(1)}{k!}(x-1)^k$, so you need $f$ and its first three derivatives, each evaluated at $1$. 2. Differentiate three times, rewriting as powers so the pattern is visible: $f(x) = \ln x$, then $f'(x) = x^{-1}$, $f''(x) = -x^{-2}$, and $f'''(x) = 2x^{-3}$. 3. Evaluate every one at the center $x = 1$: $f(1) = \ln 1 = 0$, $f'(1) = 1$, $f''(1) = -1$, and $f'''(1) = 2$. These are the raw derivative values, before any factorials. 4. Divide each by its factorial to get the coefficients: $c_0 = \frac{0}{0!} = 0$, $c_1 = \frac{1}{1!} = 1$, $c_2 = \frac{-1}{2!} = -\frac{1}{2}$, and $c_3 = \frac{2}{3!} = \frac{2}{6} = \frac{1}{3}$. 5. Attach each coefficient to its matching $(x-1)^k$ and add: $P_3(x) = 0 + 1(x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3$. 6. Drop the zero term to read off the answer: $P_3(x) = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3$. Sanity-check at the center: $P_3(1) = 0 = \ln 1$ and $P_3'(1) = 1 = f'(1)$, so the polynomial matches $f$ in value and slope at $x = 1$, exactly as designed. Answer: $P_3(x) = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3$ ### FAQ Q: What is the difference between a Taylor series and a Maclaurin series? A: A Maclaurin series is a Taylor series centered at $a = 0$, nothing more. Every Maclaurin series is a Taylor series; the name just signals that the center is $0$, so the terms are powers of $x$ and you evaluate the derivatives at $0$. When a problem centers the expansion anywhere else, $a = 1$, $a = \frac{\pi}{2}$, and so on, it is called a Taylor series, and you plug that center into $\frac{f^{(n)}(a)}{n!}$ and use $(x-a)^n$. Q: Why do you divide by $n!$ in a Taylor series? A: The factorial cancels the one that appears when you differentiate the term $c_n(x-a)^n$ exactly $n$ times: that term becomes $n!\,c_n$ at the center while every other term vanishes. To make the polynomial's $n$th derivative match $f^{(n)}(a)$, you need $n!\,c_n = f^{(n)}(a)$, which forces $c_n = \frac{f^{(n)}(a)}{n!}$. Without the factorial the higher derivatives would not line up. Q: What is the difference between a Taylor polynomial and a Taylor series? A: A Taylor polynomial stops at a chosen degree $n$; a Taylor series keeps going forever. Formally, a Taylor polynomial is a partial sum of the Taylor series (LIM-8.E.1). The builder shows a polynomial because you pick a finite degree, and as you raise that degree you are switching on more terms of the full series, tightening the fit around the center. Q: Do I have to compute every derivative to find a Taylor series? A: Often no, and recognizing when to skip it is the point. If $f$ is $e^x$, $\sin x$, $\cos x$, or $\frac{1}{1-x}$ after a substitution, constant multiple, derivative, or integral, start from that known Maclaurin series instead of differentiating repeatedly (Topic 10.15). Reserve the from-derivatives method for a nonzero center or a function that does not reduce to one of the standard series, like $\ln x$ centered at $1$. ## Derivative Practice Checker: Any Correct Form Passes URL: https://www.calclearn.org/calculators/derivative-practice-checker Last updated: 2026-07-22 Answer first: The Derivative Practice Checker shows a function, you type its derivative, and it grades by numeric equivalence, not string matching: any algebraically equal form is accepted and a constant-offset slip is always caught. A miss gets a hint naming the rule to reconsider, never the answer. The tool shows one function at a time and asks you to type its derivative into the box. Instead of comparing your typing to one stored string, it grades by numeric equivalence. Writing $g$ for the expression you type and $f'$ for the reference derivative, it evaluates both at roughly a dozen points drawn from inside the shared domain and marks you correct only when they agree at every one of them. $$\left| g(x_i) - f'(x_i) \right| < 10^{-6} \cdot \max\left(1, \left| f'(x_i) \right|\right) \quad \text{at every sampled } x_i$$ NOTE (Any correct form passes): Because the check is numeric and not textual, every algebraically equal answer counts. For $\tan x$ the tool accepts both $\sec^2 x$ and $1 + \tan^2 x$; for $\cos^2 x$ it accepts both $-2\sin x\cos x$ and $-\sin(2x)$. A quiz that stored a single answer string would reject the others, even though they are the same function. This tool cares only about the function you wrote, not how you wrote it. Sampling at many points, not one, is what makes that trust safe. An answer that is off by a constant, a dropped $+1$ say, disagrees at every point and is caught. So is a chain-rule slip that forgets the inner factor, because it changes the shape of the function, not just its value at one place. A single-point check could be fooled by either; a dozen-point check is not. - An empty box returns a gentle prompt to type something. It is never marked wrong. - Input the parser cannot read returns the parser's own message, and you are not penalized, because a typo is not a math mistake. - A parseable answer that disagrees somewhere is marked not equivalent yet, with a hint toward the rule to reconsider. - A parseable answer that agrees at every sampled point is marked correct. The ten problems are not grouped under rule headings. Before you can differentiate anything you have to read the function and decide which rule governs it, which is a graded skill in its own right: the CED calls it selecting procedures for calculating derivatives (Topic 3.5). Reading the structure first is the habit the tool is built to reward. | When the function looks like | Reach for | CED topic | | --- | --- | --- | | A sum of power terms, $x^3 - 5x + 2$ | Power rule, term by term | Topics 2.5 and 2.6 | | Two functions multiplied, $x^2\sin x$ | Product rule | Topic 2.8 | | One expression over another, $\frac{2x+1}{x^2+1}$ | Quotient rule | Topic 2.9 | | A function inside a function, $\sin(3x^2)$ | Chain rule | Topic 3.1 | | A trig function on its own, $\tan x$ | A memorized trig derivative | Topic 2.10 | | An exponential, $e^{2x}$ | Exponential rule with the chain rule | Topics 2.7 and 3.1 | | A logarithm as a factor, $x\ln x$ | The $\ln$ derivative plus the product rule | Topics 2.7 and 2.8 | These rules combine, and the tool leans on that. $x^2 e^x$ is a product of two functions you can each differentiate; $\sqrt{x^2+1}$ is a power wrapped around $x^2+1$, so it is a chain; $\cos^2 x$ means $(\cos x)^2$, a chain whose outer operation is squaring. Name the outermost structure first, then work inward, and the rule to start with becomes clear. Take the function $f(x) = \cos^2 x$. Read its structure before reaching for a rule: $\cos^2 x$ means $(\cos x)^2$, a function raised to a power, so the outer operation is squaring and the inner function is $\cos x$. That makes this a chain rule problem (Topic 3.1), not a product. $$f(x) = (\cos x)^2, \qquad \text{outer } u^2, \quad \text{inner } u = \cos x$$ Differentiate the outer square, leaving the inside untouched, then multiply by the derivative of the inside: $$f'(x) = 2\cos x \cdot (-\sin x) = -2\sin x\cos x$$ Now use the double-angle identity $\sin(2x) = 2\sin x\cos x$. The same derivative can be written $-\sin(2x)$, and both forms are exactly the same function: $$-2\sin x\cos x = -\sin(2x)$$ Type either one into the checker and it passes, because it evaluates your expression at a dozen points and both forms return identical values everywhere. Reveal the derivation afterward and the tool shows the standard form, $f'(x) = -2\sin x\cos x$, alongside the rule-by-rule steps. NOTE (How this tool teaches instead of solving): You produce the derivative and type it in; a solver would produce it for you. The tool never shows the answer until you have committed to one or asked to see the derivation. Grading is by numeric equivalence, so any algebraically correct form is accepted; a string-matching quiz would reject a right answer written a different way. A wrong answer returns a rule-scoped hint that names the rule to reconsider but never states the derivative, so the retry is your own reasoning. You decide which differentiation rule applies before differentiating, the skill graded in Topic 3.5, rather than having the method chosen for you. Because it samples about a dozen points, a constant-offset or dropped-factor mistake is always caught, unlike a check at a single point. The full rule-by-rule derivation and the standard-form answer stay behind a reveal, available only after you try, so you check your method instead of copying it. ### FAQ Q: Does the checker accept a derivative written in a different form? A: Yes. It grades by numeric equivalence, not by matching your text to a stored answer, so any algebraically equal form counts. It evaluates both your expression and the reference at about a dozen points and marks you correct when they agree everywhere. For $\tan x$ both $\sec^2 x$ and $1 + \tan^2 x$ pass; for $\cos^2 x$ both $-2\sin x\cos x$ and $-\sin(2x)$ pass. Write the derivative in whatever form your work produced. Q: I typed a correct derivative but it said it could not read the expression. Why? A: That is a grammar problem, not a wrong answer, and it is not counted against you. The parser needs functions in parentheses, as in $\sin(x)$, $\cos(x)$, or $\ln(x)$; powers written with a caret, as in x^2; and the exponential written as e^(2x) or exp(2x). One more quirk: log means natural log here. Fix the notation and check again, and the tool grades the corrected expression. Q: Will it catch a derivative that is off by a constant? A: Yes. Because the tool samples about a dozen points rather than one, an answer that differs from the reference by a constant, such as a dropped $+1$, disagrees at every point and is marked not equivalent. The same is true of a chain-rule slip that omits the inner factor, since that changes the function's shape. A single-point check could accidentally pass either mistake; this one does not. Q: Does the tool show the steps? A: Only after you have committed to an answer or asked for them. On a miss you first get a hint that names the rule to reconsider without stating the derivative, so your next attempt is still your own work. You can then reveal the full rule-by-rule derivation and the standard-form answer, which are the only place the tool ever writes out $f'(x)$. There is no score and no timer; the progress dots simply fill in as you solve each problem. Q: Which differentiation rules does the practice cover? A: The ten problems span the core AB rules: the power rule, product rule, quotient rule, and chain rule, plus the derivatives of trigonometric, exponential, and logarithmic functions. These live in Units 2 and 3 of the CED and are foundational for both AP Calculus AB and BC, so BC students use the same tool to keep the mechanics sharp. ## Related Rates Setup Builder: The Five-Step Method URL: https://www.calclearn.org/calculators/related-rates-setup-builder Last updated: 2026-07-22 Answer first: Set up a related rates problem in five steps: name the changing quantities and rates, write one equation relating them, differentiate both sides with respect to time, substitute known values only after differentiating, then solve for the unknown rate with units. You pick the governing relation yourself. This tool walks one AP-style word problem at a time through the five-step related rates method. The scenario stays pinned at the top so you can re-read it at every step, and you step through naming the quantities, choosing the governing relation, differentiating with respect to time, substituting the known values, and solving. It is a teaching tool, not a quiz: there is no score and no timer. The whole point is that at step two you have to pick the relation yourself from three candidates, and a wrong pick returns feedback on why that equation fails to model the picture without ever revealing the correct one. Related rates is CED Topic 4.4 (Introduction to Related Rates) and Topic 4.5 (Solving Related Rates Problems), tested on both AB and BC. - Name every changing quantity with a letter, then write down the rate you are given and the rate you want. - Find one equation relating those quantities: a known formula, an area or volume, or similar triangles. - Differentiate both sides with respect to time $t$, applying the chain rule to every variable. - Substitute the known values only after differentiating, never before. - Solve for the unknown rate, and state it with units and the meaning of its sign. The tool gates forward progress at step two: you cannot move on to differentiate until you have identified the correct relation. That mirrors the exam, where a wrong setup dooms every later part of a free-response question no matter how clean your algebra is. The reason step three uses implicit differentiation is that in a related rates problem every quantity is secretly a function of time. The radius of an inflating balloon, the depth of water in a tank, the distance between two cars: each one changes as $t$ advances, so each one has its own rate. When you differentiate an equation that ties those quantities together, the chain rule attaches a rate to every variable. Take the volume of a sphere as an example, where both $V$ and $r$ depend on $t$: $$V = \frac{4}{3}\pi r^3 \quad\Longrightarrow\quad \frac{dV}{dt} = 4\pi r^2 \,\frac{dr}{dt}$$ The factor $\frac{dr}{dt}$ appears because $r$ is a function of $t$, so differentiating $r^3$ with respect to $t$ is $3r^2$ times the inner rate $\frac{dr}{dt}$. That extra rate factor is the whole difference between an ordinary derivative and a related rate. This is why related rates uses Leibniz notation such as $\frac{dV}{dt}$ and $\frac{dr}{dt}$ rather than prime notation: you need to name which variable is changing and with respect to what, and here it is always with respect to time. A compute engine will happily differentiate whatever equation you hand it, but it will never tell you which equation to write down in the first place. That recognition is the actual work of a related rates problem, and it is what the builder forces you to practice. Every relation you would ever pick comes from a short menu of geometric facts, and the shape of the scenario tells you which one applies. | Scenario in the problem | Relation to write | Why | | --- | --- | --- | | A right triangle: ladder on a wall, two paths at 90 degrees | $x^2 + y^2 = c^2$ | The Pythagorean theorem ties the two legs to the hypotenuse; the legs never simply add. | | A sphere growing or shrinking | $V = \frac{4}{3}\pi r^3$ | The material fills the interior, so it is the volume, not the surface area, that is tied to the radius. | | A circle spreading, like a ripple | $A = \pi r^2$ | The enclosed region is a flat disk, so its size is the area of a circle, not its circumference. | | A cone filling with liquid | $V = \frac{1}{3}\pi r^2 h$, with $r$ from similar triangles | The surface radius grows with depth, so replace $r$ using the tank's fixed proportions before differentiating. | | A shadow cast by a light above | $\frac{h_1}{s} = \frac{h_2}{x+s}$ | The object and the pole form similar triangles that share the same light ray to the shadow tip. | NOTE (A wrong relation leads nowhere): In the builder, picking the surface area $4\pi r^2$ for the inflating balloon, or the circumference $2\pi r$ for the spreading ripple, returns feedback naming exactly why that formula does not model the scenario. The correct choice is never revealed, so a retry is reasoning, not copying. Differentiating the wrong equation produces a tidy-looking but meaningless answer, which is the trap a solver cannot warn you about. The single most expensive mistake in related rates is substituting a number for a changing quantity before you differentiate. It is tempting because the problem hands you a value, but that value is only true at one instant. Freeze it into the equation and its derivative becomes zero, which is a false statement about a quantity that is plainly still moving. Differentiate while the letters are still variables, and only then put the numbers in. $$\frac{d}{dt}\left(x^2 + y^2\right) = \frac{d}{dt}\left(100\right) \quad\Longrightarrow\quad 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0$$ For the sliding ladder, $100$ on the right is a genuine constant (the ladder length never changes), so its derivative really is zero. But $x = 6$ is not a constant; it is the base distance at this one instant. If you had plugged $x = 6$ in first, you would have differentiated $36 + y^2 = 100$ and lost the $\frac{dx}{dt}$ term entirely. The builder flags this order-of-operations trap on exactly the scenarios that tempt it, the ladder and the cone. NOTE (The cone needs one more move before you differentiate): When a cone fills, both the surface radius $r$ and the depth $h$ change, so $V = \frac{1}{3}\pi r^2 h$ has two changing variables. Use the tank's fixed proportions to write $r$ in terms of $h$ first. A tank $10$ ft tall with a $5$ ft top radius gives $\frac{r}{h} = \frac{5}{10}$, so $r = \frac{h}{2}$ and $V = \frac{\pi}{12}h^3$. Reducing to one variable before differentiating is what makes $\frac{dh}{dt}$ solvable. A 10-foot ladder leans against a vertical wall. The base slides away from the wall at a constant 2 feet per second. How fast is the top sliding down the wall at the instant the base is 6 feet from the wall? Here is the whole method the builder walks, start to finish. Step 1, name the quantities. Let $x$ be the distance from the wall to the base and $y$ the height of the top on the wall. You are given $\frac{dx}{dt} = 2$ and you want $\frac{dy}{dt}$ at the instant $x = 6$. Step 2, choose the relation. The wall, the ground, and the ladder form a right triangle whose hypotenuse is the fixed 10-foot ladder, so the Pythagorean theorem ties the pieces together: $$x^2 + y^2 = 10^2 = 100$$ Step 3, differentiate with respect to time. Both $x$ and $y$ are functions of $t$, so each picks up its own rate by the chain rule, while the constant on the right differentiates to zero: $$2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0$$ Step 4, substitute the known values, now that differentiating is done. At this instant $x = 6$ and $\frac{dx}{dt} = 2$. The height $y$ is not given, so recover it from the relation: $y = \sqrt{100 - 6^2} = \sqrt{64} = 8$. Step 5, solve for the unknown rate and interpret. Isolate $\frac{dy}{dt}$, then substitute: $$\frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt} = -\frac{6}{8}\cdot 2 = -\frac{3}{2} = -1.5 \ \text{ft/s}$$ The top of the ladder slides downward at 1.5 ft/s. The negative sign is not a mistake to drop: it is the answer telling you the height $y$ is decreasing while the base moves out. Reading the rate back with its units and the meaning of its sign is step five, and it is where a lot of otherwise-correct work loses its last point. NOTE (How this tool teaches instead of solving): You pick the governing equation yourself from three candidates, so you practice the recognition that word problems actually test; a solver picks the relation silently and hands you only a number. Wrong relation choices return scenario-specific feedback explaining why that equation fails to model the picture, without revealing the correct one, so a retry is reasoning rather than copying. The tool gates progress at the relation step: you cannot advance to differentiate until you identify the correct relation, mirroring how a wrong setup dooms the rest of an AP free-response part. Each step is kept separate on screen, so you see naming, implicit differentiation with respect to time, substitution, and solving as distinct moves instead of one collapsed answer. On the ladder and cone scenarios it flags the substitute-only-after-differentiating trap explicitly, the mistake that freezes a changing quantity into a constant with derivative zero. Every final answer is stated with units and the meaning of its sign, so you practice interpreting the rate, not just computing it. ### FAQ Q: What are the five steps to set up a related rates problem? A: First, name every changing quantity and write down the rate you are given and the rate you want. Second, find one equation relating those quantities, whether a formula, an area or volume, or similar triangles. Third, differentiate both sides with respect to time $t$, applying the chain rule to every variable. Fourth, substitute the known values, but only after differentiating. Fifth, solve for the unknown rate and state it with units and the meaning of its sign. The builder walks each step in that order and will not let you skip the relation. Q: Why do I substitute the numbers only after differentiating? A: Because the numbers a related rates problem gives you are true at only one instant, not for all time. If you plug $x = 6$ into $x^2 + y^2 = 100$ before differentiating, you freeze $x$ into a constant, its derivative becomes zero, and you lose the $\frac{dx}{dt}$ term you needed. Differentiate while $x$ and $y$ are still variables, so the chain rule attaches a rate to each, and only then substitute the values for this instant. The cone problem adds a wrinkle: reduce to one variable with similar triangles before differentiating, since the surface radius changes too. Q: How do I know which equation to use? A: Read the geometry of the scenario. A right triangle, like a ladder on a wall or two paths meeting at 90 degrees, calls for the Pythagorean theorem $x^2 + y^2 = c^2$. A growing sphere uses the volume $V = \frac{4}{3}\pi r^3$, a spreading circle uses the area $A = \pi r^2$, and a filling cone uses $V = \frac{1}{3}\pi r^2 h$ with the radius rewritten from the tank's proportions. A shadow problem uses similar triangles. The builder's step two is exactly this recognition, and its feedback on a wrong pick names why that formula does not fit. Q: What is the difference between related rates and implicit differentiation? A: They are the same technique aimed at different variables. Implicit differentiation (Topic 3.2) differentiates a relation with respect to $x$, treating $y$ as a function of $x$. Related rates (Topics 4.4 and 4.5) differentiates a relation with respect to time $t$, treating every quantity as a function of $t$. In both cases the chain rule is what attaches the extra factor, $\frac{dy}{dx}$ in one case and rates like $\frac{dV}{dt}$ in the other. If you are comfortable with implicit differentiation, related rates is that same move with $t$ as the driving variable. Q: Are related rates on the AP Calculus exam? A: Yes, on both AB and BC. They are Topic 4.4 (Introduction to Related Rates) and Topic 4.5 (Solving Related Rates Problems) in Unit 4, Contextual Applications of Differentiation. They appear on the exam, and where a related rates problem is scored, credit is given for the setup, choosing the relation and differentiating with respect to time, not only the final number. Practicing the setup is exactly what this tool is built for. ## Optimization Setup Builder: Objective vs Constraint URL: https://www.calclearn.org/calculators/optimization-setup-builder Last updated: 2026-07-22 Answer first: Set up an optimization problem by deciding what you are optimizing (the objective) and separating it from what the problem holds fixed (the constraint). Use the constraint to rewrite the objective in one variable, state the domain, differentiate, find critical points, and confirm the max or min with a derivative test. The hardest part of an optimization problem is not the calculus, it is deciding what to optimize. This builder makes you commit to that decision first: it shows a scenario and three candidate formulas, and you choose the one quantity the problem asks you to maximize or minimize. The other two are things you should not optimize: usually a quantity the problem holds fixed (a constraint), and sometimes a related quantity that is the wrong thing to make large or small. Getting this split right is the heart of Topics 5.10 and 5.11 (Introduction to Optimization Problems and Solving Optimization Problems); almost everything after it is procedure. The objective is the quantity that is free to change and that you want as large or as small as possible: an area, a volume, a length of fence, a distance. The constraint is whatever the problem pins to a fixed value: the $100$ meters of fencing you already own, the $1800$ square feet the pen must enclose, the $12$ inch sheet you start with. A quick test settles most cases: if the problem states a quantity as a specific number, that number is the constraint, so it cannot be the thing you optimize. NOTE (The mistake the tool is built to catch): Students routinely set up to optimize the quantity that is actually fixed. If you try to maximize the perimeter when the fencing is already $100$ meters, there is nothing to maximize, the perimeter is stuck at $100$. When you pick one of the other two candidates in the builder, it explains specifically why that quantity is not the objective, often because the problem holds it fixed, and lets you try again; it never names the correct choice, so the retry is reasoning rather than copying. Once you have named the objective, the builder reveals the rest of the setup one line at a time, in the order the AP rubric rewards. You write the constraint equation, use it to rewrite the objective in a single variable, state the domain, differentiate, solve for the critical points, and run a derivative test before you answer. A persistent checklist keeps the full procedure in view so you internalize the method, not one answer. - Draw and label a picture. - Name the variables. - Write the objective, the quantity you optimize. - Write the constraint, the quantity that is fixed. - Substitute to get the objective in one variable. - State the domain, endpoints included. - Differentiate and find the critical points. - Run a derivative test, then answer the actual question with units. The reason for reducing to one variable is that the derivative tools you know act on a function of a single input. The constraint is what lets you do it: solve the constraint for one variable and substitute, and a two variable objective collapses into one you can differentiate. For a rectangle of fixed perimeter, the constraint $2x + 2y = 100$ gives $y = 50 - x$, and the area $A = xy$ becomes a function of $x$ alone. $$2x + 2y = 100 \;\Rightarrow\; y = 50 - x \;\Rightarrow\; A(x) = x(50 - x) = 50x - x^2$$ With the objective in one variable you differentiate, set the derivative equal to $0$ to find the critical points, then confirm which one is the maximum or minimum. Skipping that confirmation is the most common way to lose points: a critical point is only a candidate until a derivative test, or a comparison against the endpoints, proves it wins. One step in the checklist does more work than it looks: state the domain, endpoints included. The physical setup limits how large or small your variable can be, and those limits matter, because on a closed interval the maximum or minimum can occur at an endpoint rather than at an interior critical point. This is the candidates test for absolute extrema (Topic 5.5): on a closed interval, compare the objective at every critical point and at both endpoints, and the largest or smallest value wins. For the open-top box cut from a $12$ inch square, the corner cut $x$ has to satisfy $0 \le x \le 6$: a cut of $0$ leaves a flat sheet, and a cut of $6$ removes the whole side. At both endpoints the volume is $0$, so neither endpoint can be the maximum, which is why the interior critical point governs here. State the domain anyway, every time, because the moment the endpoints are not obvious losers you will need them. NOTE (Closed interval versus open): On a closed interval like $0 \le x \le 6$ the endpoints are fair game and must be checked. On an open interval like $x > 0$, the three-sided-pen (minimum-fence) and cylinder problems for instance, no endpoint is attainable, so an interior critical point that a derivative test confirms is the answer. The maximum-area fencing problem, by contrast, is closed, $0 \le x \le 50$, so its endpoints belong in the check. Reading the domain tells you which situation you are in before you commit to an answer. ### Worked example #### The largest open-top box Problem: An open-top box is formed from a $12$ inch by $12$ inch square sheet by cutting a square of side $x$ from each corner and folding up the flaps. Find the value of $x$ that gives the largest volume. 1. Step 1, name the objective. You are asked for the largest volume, so the objective is the volume $V$. The area of the original sheet is fixed at $144$ square inches, so it is a constraint, not something you can maximize; the base area alone is largest when $x = 0$, but then the box has no height and no volume, so that is not the goal either. 2. Step 2, write the constraint and reduce to one variable. Cutting $x$ from each corner leaves a base of side $12 - 2x$ and a height of $x$, so the volume is the product of these: $V(x) = x(12 - 2x)^2$. 3. Step 3, state the domain. The cut cannot be negative and cannot exceed half the side, so $0 \le x \le 6$. At $x = 0$ and $x = 6$ the volume is $0$, so if there is a positive maximum it lives strictly inside the interval. 4. Step 4, differentiate. Using the product and chain rules and factoring, the derivative comes out cleanly: $V'(x) = (12 - 2x)(12 - 6x)$. 5. Step 5, find the critical points. Set $V'(x) = 0$: $(12 - 2x)(12 - 6x) = 0 \;\Rightarrow\; x = 6 \text{ or } x = 2$. The solution $x = 6$ is an endpoint where the volume is $0$, so discard it. The candidate is $x = 2$. 6. Step 6, run a derivative test. Check the sign of $V'$ on either side of $x = 2$: $V'(1) = (10)(6) = 60 > 0$ and $V'(3) = (6)(-6) = -36 < 0$. The derivative changes from positive to negative, so $V$ has a maximum at $x = 2$ by the first derivative test. 7. Step 7, answer the actual question with units. The problem asked for the cut length that maximizes volume, so report $x = 2$ inches, and evaluate the volume there as a check: $V(2) = 2(12 - 4)^2 = 2 \cdot 64 = 128$. Answer: A corner cut of $x = 2$ inches gives the largest box, with volume $128$ cubic inches. Notice where the last two steps land: the derivative test justifies the maximum, and the answer carries units. On a free response question, both are where the points live. ### FAQ Q: How do I tell the objective from the constraint in an optimization problem? A: The objective is the quantity you are asked to make as large or as small as possible, and it is free to change: an area, a volume, a length, a distance. The constraint is whatever the problem fixes to a specific value, like $100$ meters of fencing or an enclosed area of $1800$ square feet. A fast test: if the problem states a quantity as a definite number, that number is the constraint, so it cannot be the thing you optimize. Name the objective first, then use the constraint to reduce it to one variable. Q: Why do I have to state the domain before I optimize? A: Because the maximum or minimum can sit at an endpoint, not at an interior critical point. On a closed interval the candidates test (Topic 5.5) says to compare the objective at every critical point and at both endpoints and take the winner. For the open-top box the domain is $0 \le x \le 6$, and the volume is $0$ at both endpoints, so the interior critical point wins; but you only know that after you have written the domain down. When the endpoints are not obvious losers, forgetting them costs the answer. Q: Should I use the first derivative test or the second derivative test? A: Either one justifies a maximum or minimum, so use whichever is less work for your function. The first derivative test checks the sign of $f'$ just before and just after the critical point: a change from positive to negative is a maximum, negative to positive is a minimum. The second derivative test checks concavity: $f'' < 0$ at the critical point gives a maximum, $f'' > 0$ gives a minimum. The second test is quick when $f''$ is easy to evaluate, but it is inconclusive when $f'' = 0$, so fall back to the first derivative test or the candidates test there. Q: Is optimization on the AP Calculus exam? A: Yes, in both AB and BC, as Topics 5.10 and 5.11. It appears in multiple choice and, classically, as a free response question that asks you to build the model and justify the extreme value. The graded steps are the setup and the justification: defining the objective, reducing to one variable, and confirming the maximum or minimum with a derivative test or the candidates test. Reporting a critical point without that justification does not earn full credit, which is exactly why this tool drills the setup rather than the number. ## Interval of Convergence Finder: Ratio Test and Endpoints URL: https://www.calclearn.org/calculators/interval-of-convergence-finder Last updated: 2026-07-22 Answer first: Find an interval of convergence in two stages: run the ratio test and solve L < 1 for the radius R and an open interval centered at c, then test each endpoint separately, because the ratio test gives L = 1 there and settles nothing. The final interval can be open, half-open, or closed. This tool takes one power series at a time and walks it through the ratio test in five staged reveals, in the order you would write on the exam: assemble the term ratio, take its limit $L$, solve $L < 1$ for the radius, test the endpoints, then draw the interval. Finding the radius and interval of convergence of a power series is Topic 10.13, and the engine that produces the radius is the ratio test from Topic 10.8. For a power series $\sum a_n$ with terms $a_n = c_n (x - c)^n$, the ratio test compares the size of one term to the last. The powers of $(x - c)$ pull a factor of $|x - c|$ out front, and whatever is left from the coefficients $c_n$ collapses to a number in the limit: $$L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|, \qquad \text{the series converges when } L < 1$$ Because $L$ always comes out proportional to $|x - c|$, the condition $L < 1$ rearranges into a distance from the center. That distance is the radius $R$: $$L < 1 \iff |x - c| < R$$ Solving that inequality gives the open interval $(c - R,\, c + R)$. The tool labels this the open interval so far, because the two endpoints $x = c - R$ and $x = c + R$ are still undecided. The ratio test is deliberately silent at the endpoints. Substitute $x = c \pm R$ and the distance $|x - c|$ equals $R$ exactly, so $L = 1$, which is the one case the ratio test cannot settle (Topic 10.8): $$x = c - R \;\text{or}\; x = c + R \;\Rightarrow\; L = 1 \quad (\text{inconclusive})$$ So the tool does the step a solver skips: it substitutes each endpoint value back into the series to get a concrete number series, then picks a named test for it. An endpoint that produces an alternating series usually converges by the Alternating Series Test (Topic 10.7); one that produces the harmonic series $\sum \frac{1}{n}$ diverges as a $p$-series with $p = 1$ (Topic 10.5); and terms that do not shrink to $0$ fail the nth-term test (Topic 10.3). Because each endpoint is decided on its own, the interval can end up open, half-open, or closed. NOTE (One endpoint can converge while the other diverges): That is why the number line uses two kinds of dots. A filled dot means the endpoint converges and is included (a square bracket); an open dot means it diverges and is excluded (a parenthesis). The series $\sum \frac{x^n}{n}$ converges at $x = -1$ (alternating) but diverges at $x = 1$ (harmonic), so its interval is the half-open $[-1,\, 1)$: one bracket, one parenthesis. Take the series the tool centers away from the origin, $\sum_{n=1}^{\infty} \frac{(x-2)^n}{n \cdot 3^n}$. Its center is $c = 2$, the detail most students drop. Watch the radius come from the ratio test and the interval come from the endpoints. First assemble the term ratio and simplify: $$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x-2)^{n+1}}{(n+1) \cdot 3^{n+1}} \cdot \frac{n \cdot 3^n}{(x-2)^n}\right| = \frac{|x-2|}{3} \cdot \frac{n}{n+1}$$ Take the limit as $n \to \infty$; the fraction $\frac{n}{n+1}$ tends to $1$, leaving $L$ proportional to $|x - 2|$: $$L = \lim_{n \to \infty} \frac{|x-2|}{3} \cdot \frac{n}{n+1} = \frac{|x-2|}{3}$$ Set the limit below $1$ and read off the radius: $$L < 1 \iff |x-2| < 3 \;\Rightarrow\; R = 3$$ A radius of $3$ about the center $c = 2$ gives the open interval $(2 - 3,\, 2 + 3) = (-1,\, 5)$. Now test both endpoints by substituting each $x$-value into the series: $$x = -1: \quad \sum_{n=1}^{\infty} \frac{(-3)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \quad (\text{converges})$$ $$x = 5: \quad \sum_{n=1}^{\infty} \frac{3^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{1}{n} \quad (\text{diverges})$$ At $x = -1$ the series is the alternating harmonic series, which converges by the Alternating Series Test, so the left endpoint is included. At $x = 5$ the series is the plain harmonic series, a $p$-series with $p = 1$, which diverges, so the right endpoint is excluded. One bracket, one parenthesis, giving the interval of convergence: $$[-1,\, 5)$$ The radius $R$ can be a finite positive number, infinity, or zero, and the tool's five series include at least one of each kind (three with a finite radius, one with $R = \infty$, one with $R = 0$), so you learn to read the shape from the terms. The number line renders all three: a segment between two endpoints, the whole line with arrowheads, or a single point at the center. | Radius | Converges for | Endpoints | Series in the tool | | --- | --- | --- | --- | | $R$ finite, $R > 0$ | $x$ within $R$ of the center | test both, separately | $\sum \frac{x^n}{n}$ gives $[-1,\, 1)$ | | $R = \infty$ | every real number | none to test | $\sum \frac{x^n}{n!}$, the series for $e^x$ | | $R = 0$ | only the center $x = c$ | the interval is one point | $\sum n! \cdot x^n$ converges only at $0$ | NOTE (When only odd powers appear): The series $\sum \frac{(-1)^n x^{2n+1}}{2n+1}$, the Maclaurin series for $\arctan x$, has only odd powers, so the consecutive-term ratio tends to $x^2$, not $|x|$; solving $x^2 < 1$ still gives $R = 1$. Both endpoints turn into alternating series that converge, so the interval is the closed $[-1,\, 1]$, and $x = 1$ recovers the Leibniz formula for $\frac{\pi}{4}$. NOTE (How this tool teaches instead of solving): You assemble the term ratio and simplify it to the limit $L$ yourself; a compute-only solver just prints the radius. You solve $L < 1$ for the radius $R$ and write the open interval before deciding any endpoint, so the two-stage structure stays in front of you. You substitute each endpoint into the series and choose its own named test, which is the exact step that makes the interval open, closed, or half-open. The number line marks included endpoints with filled dots and excluded ones with open dots, so a bracket versus a parenthesis becomes something you can see. Five series cover the finite, infinite, and zero-radius cases, training you to recognize the regime instead of memorizing one interval. ### FAQ Q: What is the difference between the radius and the interval of convergence? A: The radius of convergence $R$ is a single distance: the series converges for every $x$ within $R$ of the center $c$ and diverges beyond it. The interval of convergence is the actual set of $x$ values, and getting it takes one more step, testing the two endpoints $x = c \pm R$. The radius tells you the interval is $(c - R,\, c + R)$ up to its edges; the endpoint tests decide whether each edge is included, so the interval can be open, half-open, or closed (Topic 10.13). Q: Why can't the ratio test decide the endpoints? A: At an endpoint $x = c \pm R$ the distance $|x - c|$ equals the radius $R$ exactly, so the ratio test limit comes out to $L = 1$. That is the inconclusive case of the ratio test (Topic 10.8): $L < 1$ proves convergence and $L > 1$ proves divergence, but $L = 1$ proves nothing. So you substitute the endpoint value into the series to get a number series and apply a different test, usually the Alternating Series Test or the $p$-series rule. Q: Can the interval of convergence be a single point or all real numbers? A: Yes, both extremes happen. If $R = \infty$ the series converges for every real number and the interval is $(-\infty,\, \infty)$; the series $\sum \frac{x^n}{n!}$ for $e^x$ is the standard example. If $R = 0$ the series converges only at its center and the interval collapses to the single point $\{c\}$; $\sum n! \cdot x^n$ does this, because the factorial forces the terms to blow up for any $x \ne 0$. Every power series converges at least at its center, so the interval is never empty. Q: How do I know whether to use a bracket or a parenthesis at an endpoint? A: Substitute the endpoint's $x$-value into the power series, which turns it into an ordinary number series, then test that series. If it converges, the endpoint is included and you use a square bracket; if it diverges, the endpoint is excluded and you use a parenthesis. For $\sum \frac{x^n}{n}$, the endpoint $x = -1$ gives the convergent alternating harmonic series (bracket), while $x = 1$ gives the divergent harmonic series (parenthesis), so the interval is $[-1,\, 1)$. Q: Is finding the interval of convergence on the AP Calculus BC exam? A: Yes. It is Topic 10.13 in Unit 10, Infinite Sequences and Series, which is BC-only and weighted 15 to 20 percent of the BC exam. Expect to find a radius with the ratio test and then justify each endpoint with a named test; the endpoint work is where most of the points live, since the ratio test alone only gets you the open interval. --- # Interactive visualizers ## Tangent Line Tracer: Watch the Derivative Appear URL: https://www.calclearn.org/interactives/tangent-line-tracer Last updated: 2026-07-22 Answer first: A derivative is the slope of the tangent line to a curve at each point. Slide a point along f(x), read how steep the tangent is there, and plot that steepness as a height. String those heights together and they trace a whole new curve, f'(x): the slope of f, recorded everywhere at once. A derivative answers one question at every point on a graph: how steep is the curve right here? Pick a point on $f$, draw the line that just grazes the curve there (the tangent line), and measure its slope. That slope is the value of $f'$ at that point. Topic 2.2 states it in one line: the derivative of a function at a point is the slope of the line tangent to the graph at that point (CHA-2.C.1). $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ That limit hands back a single number for a single input. The move that gives calculus its reach is running it at every $x$ at once. Slide the point across $f$, record the tangent slope at each location, and the recorded slopes stack into a graph of their own. The height of that new graph at any $x$ equals the steepness of $f$ at the same $x$. The new graph is the function $f'(x)$, and reading it against $f$ is the whole skill of Topics 5.8 and 5.9. NOTE (Read the derivative off the shape, before any formula): You can sketch the sign of $f'$ without differentiating anything. Where $f$ climbs, its tangent tilts up, so $f' > 0$ and the derivative curve sits above the axis. Where $f$ falls, the tangent tilts down, so $f' < 0$ and the curve drops below. Where $f$ levels off at a peak or a valley, the tangent is flat, so $f'$ touches zero and switches sign. That is exactly Topic 5.3 (the sign of $f'$ marks where $f$ increases and decreases) and Topic 5.2 (a zero of $f'$ is a critical point, a candidate for a maximum or minimum). Steepness carries size as well as sign. A near-vertical stretch of $f$ throws $f'$ far from the axis; a lazy, gently sloping stretch keeps $f'$ hugging it. Watching the two graphs move in step turns the derivative from a rule you memorize into a shape you can predict before you compute a thing. ### How to read it The screen holds two linked panels. On the left is $f(x)$ with a point you can grab and a tangent line pinned to it. On the right is a blank pair of axes that fills in as you move: every position of the point drops a dot at a height equal to the current tangent slope. Drag slowly and the right panel draws the graph of $f'(x)$ one dot at a time. - Drag onto a steep uphill section and watch the right-panel dot ride high above the axis; the steeper the climb, the higher it sits. - Drag onto a downhill section. The tangent tilts the other way and the dot falls below the axis into negative slope. - Ease the point onto a peak or a valley of $f$. The tangent goes flat and the derivative dot lands right on the axis at $f'(x) = 0$. - Sweep the point all the way across, let the full curve of $f'$ appear, then check its shape against the prediction you made from $f$ alone. NOTE (The tell for a maximum): Watch the order of events at a hilltop of $f$. Just before the peak the tangent still tilts up, so $f' > 0$; at the peak it is flat, so $f' = 0$; just after, it tilts down, so $f' < 0$. The derivative curve crossing from positive to negative is the graphical form of the First Derivative Test (Topic 5.4). A valley shows the mirror image: $f'$ crosses from negative to positive. If the tracer lets you switch functions, try one with a sharp corner such as $f(x) = |x|$. Approach the corner from the left and the right, and the tangent slope disagrees: no single value of $f'$ exists there, so the right panel jumps rather than drawing a smooth point. That is Topic 2.4 in the open: a continuous graph can still fail to be differentiable, because the one-sided difference quotients do not meet (FUN-2.A.2). ### FAQ Q: Is the derivative a number or a function? A: Both, depending on what you ask for. $f'(a)$ at a single input is one number: the slope of the tangent at $x = a$. Collect that slope at every input and you get $f'(x)$, a function in its own right (Topic 2.2). The tracer shows the switch happening in front of you: one dot is a number, the whole traced curve is the function. Q: Why does the derivative equal the slope of the tangent line? A: The difference quotient shown above is the slope of a secant line through two points on $f$. As $h \to 0$ the second point slides into the first and the secant pivots into the tangent. The limit of those secant slopes is the tangent slope, which is exactly the definition of $f'(x)$ (Topics 2.1-2.2). Q: How do I sketch $f'$ from a graph of $f$ without a formula? A: Walk across $f$ from left to right and read its slope. Where $f$ rises, plot $f'$ above the axis; where it falls, plot $f'$ below; at every peak, valley, or flat spot, plot $f' = 0$. Steeper parts of $f$ push $f'$ farther from the axis, gentler parts pull it closer. Mark the zeros first, then fill in between them. That is the graph-reading skill tested in Topic 5.8. ## Riemann Sum Slider: Drag n, Watch the Error Vanish URL: https://www.calclearn.org/interactives/riemann-sum-slider Last updated: 2026-07-22 Answer first: A Riemann sum estimates area under a curve by slicing the interval into n rectangles and summing their areas; the definite integral is the value that sum approaches as n grows. Drag n from 1 to 200 and the rectangle tops collapse onto the curve as the error falls toward zero. Midpoint and trapezoid converge fastest. This slider turns the definition of the definite integral into something you can watch happen. Cut the interval $[a,b]$ into $n$ equal strips, cap each strip with a rectangle, and add their areas: that total is a Riemann sum, an estimate of the area under the curve (Topics 6.2 and 6.3). When $f$ is a rate of change, that area is the change it accumulates (CHA-4.A.1). Drag $n$ upward and every strip narrows, its top has less room to miss the curve, and the running total creeps toward one fixed number. $$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^{*})\,\Delta x, \qquad \Delta x = \frac{b-a}{n}$$ That fixed number is the definite integral. As the strip widths shrink toward $0$, the Riemann sums approach $\int_a^b f(x)\,dx$ exactly, and this limit is the definition of the integral (LIM-5.C.1). The error readout is the distance between your current sum and that exact value, so dragging $n$ from $1$ toward $200$ is a hands-on version of taking the limit. The left, right, and midpoint buttons change one thing only: the sample point $x_i^{*}$, meaning where each strip reads its single height (the left edge, the right edge, or the center). The trapezoidal button instead reads both edges of each strip and slants the top between them, which is the same as averaging the left and right sums. Keep $n$ fixed and click through all four to see which tops overshoot the curve and which fall short. NOTE (Predict which side you land on before you compute): You can name which side of the true value an estimate falls on with no arithmetic, and this recognition is worth quick points on multiple choice (Topic 6.2, LIM-5.A.4). Monotonicity settles the one-sided sums: for an increasing $f$ the left sum sits under the curve and the right sum sits above, and a decreasing $f$ flips both. Concavity settles the other two: on a concave-up graph the midpoint sum reads low and the trapezoidal sum reads high, trapping the exact integral between them, and concave down flips both. ### How to read it The panel shows the graph of $f$ on $[a,b]$ with the current strips shaded under it, the value of the Riemann sum, the exact integral, and the error between them. The slider along the bottom sets $n$, the number of subintervals, from $1$ up to $200$. The four buttons pick left, right, midpoint, or trapezoidal, and the strips redraw the instant you change either control. - Drag $n$ from $1$ to $200$ on any method and watch the error fall toward $0$ while the strips fuse into the shaded region under the curve. - Pick an increasing stretch and compare left versus right at the same $n$: the left tops tuck under the curve, the right tops poke above, and the true value sits bracketed between the two sums. - Put midpoint and trapezoidal head to head against left and right at the same $n$: the centered and slanted tops post a smaller error right away. - Push $n$ to $200$ and note how many decimal places of the sum now match the exact value. Freeze the slider at $n = 4$ with $f(x) = x^2$ on $[0, 2]$ and check the left rule by hand. Here $\Delta x = \frac{2}{4} = 0.5$, the left endpoints are $0, 0.5, 1, 1.5$, and their heights are $0, 0.25, 1, 2.25$. The heights sum to $3.5$, and multiplying by $\Delta x = 0.5$ gives $1.75$. The exact area is the definite integral: $$\int_0^2 x^2\,dx = \frac{8}{3} \approx 2.667$$ The left estimate of $1.75$ undershoots that true value by about $0.9$. Switch methods at the same $n$ and each one reads the height differently, so each lands somewhere else: | Rule | Reads height at | Estimate at $n = 4$ | Side of $\frac{8}{3} \approx 2.667$ | | --- | --- | --- | --- | | Left | left endpoints | $1.75$ | under | | Right | right endpoints | $3.75$ | over | | Midpoint | midpoints | $2.625$ | under (close) | | Trapezoidal | edge average | $2.75$ | over (close) | The other three rules keep the same $\Delta x = 0.5$ but read the strip heights elsewhere. Right: the right endpoints $0.5, 1, 1.5, 2$ have heights $0.25, 1, 2.25, 4$, which sum to $7.5$, and $7.5 \cdot 0.5 = 3.75$. Midpoint: the midpoints $0.25, 0.75, 1.25, 1.75$ have heights $0.0625, 0.5625, 1.5625, 3.0625$, which sum to $5.25$, and $5.25 \cdot 0.5 = 2.625$. Trapezoidal: average the left and right totals, $(1.75 + 3.75)/2 = 2.75$. NOTE (Watch how fast the error falls, not just that it falls): Double $n$ on a left or right sum and the error roughly halves; double it on a midpoint or trapezoidal sum and the error drops to roughly a quarter. That gap is the whole reason free-response questions that hand you a table of values ask for a trapezoidal estimate: with only a few data points, a slanted top wrings far more accuracy out of the same strips. The exam never asks you to compute a Riemann sum error bound, but the slider makes the pattern impossible to miss. ### FAQ Q: Does the left or right Riemann sum overestimate? A: It depends on whether the function is increasing or decreasing, not on the words left or right. On an increasing interval the right sum overestimates and the left sum underestimates; on a decreasing interval it is reversed. The slider makes it obvious: pick an increasing stretch and every right rectangle pokes above the curve while every left rectangle tucks below it. Q: Which Riemann sum is the most accurate? A: For the same $n$ the midpoint and trapezoidal sums are almost always closer than the left or right sums, because a centered or slanted top follows a sloping curve better than a top pinned to one edge. Drag $n$ up with each method selected and the error readout confirms it: left and right trail behind while midpoint and trapezoidal race to the exact value. One exception is a straight-line $f$, where the midpoint and trapezoidal sums are exact at every $n$. Q: What happens to a Riemann sum as n approaches infinity? A: As $n \to \infty$ the subinterval widths shrink to zero and the Riemann sum approaches one number, the definite integral $\int_a^b f(x)\,dx$. That limit is the formal definition of the definite integral (Topic 6.3, LIM-5.C.1). On the slider you see it as the error collapsing toward zero as you drag $n$ toward $200$, though for a curved graph no finite $n$ ever drives the error to exactly zero. ## Secant to Tangent: The Limit Definition, Made Physical URL: https://www.calclearn.org/interactives/secant-to-tangent Last updated: 2026-07-22 Answer first: The derivative f'(a) is the slope of the tangent line at x=a. You get it from the slope of a secant line through two nearby points: as h approaches 0, the difference quotient (f(a+h)-f(a))/h stops being an average rate of change and becomes the instantaneous rate at the point. A single point on a graph has no rate of change. Speed, slope, and rate all compare two things, so at one isolated instant there is nothing to compare. The limit definition of the derivative works around this: start with two points, measure the average rate of change between them, then slide the second point toward the first until the gap vanishes. The number that the average rate settles on is $f'(a)$, the instantaneous rate of change at $x = a$ (Topics 2.1 and 2.2). $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$ Geometrically, $\frac{f(a+h) - f(a)}{h}$ is the slope of the secant line through the two points $(a, f(a))$ and $(a+h, f(a+h))$. Shrinking $h$ walks the second point along the curve toward the first, and the secant line pivots. In the limit the two points merge and the secant becomes the tangent line at $x = a$. Average rate of change over the interval between $a$ and $a+h$ turns into instantaneous rate of change at the single point $a$. NOTE (Method check: is this limit secretly a derivative?): The most tested payoff of this definition is running it in reverse. If you meet a limit like $\lim_{h \to 0} \frac{(2+h)^3 - 8}{h}$ and it looks like a dead end, check whether it matches the difference quotient. Here $(2+h)^3$ matches $f(a+h)$ with $f(x) = x^3$ and $a = 2$. Check the constant: $f(a) = 2^3 = 8$ matches the $-8$ in the numerator, so the whole limit is just $f'(2) = 3(2)^2 = 12$. Recognizing the form lets you skip the algebra and differentiate instead (Topic 2.7, LIM-3.A.1). Notice you can never simply set $h = 0$. Substituting gives $\frac{f(a) - f(a)}{0} = \frac{0}{0}$, which is undefined: the difference quotient does not exist exactly at the point. That indeterminate form is the whole reason calculus needs limits. The limit lets the ratio have a value that the expression itself never reaches (Topic 1.1). ### How to read it The screen shows a curve $y = f(x)$ with a fixed anchor point at $x = a$ and a second point at $x = a + h$ that you control. A straight secant line runs through both points, and a readout reports the current value of $h$ and the current difference quotient (the secant slope). - Drag $h$ toward 0. Watch the movable point slide down the curve toward the anchor and the secant line pivot as it goes. - Keep your eye on the difference-quotient readout. As $h$ shrinks, the number stops wandering and homes in on one value: that value is $f'(a)$, the slope of the tangent. - Notice the secant never needs $h$ to actually hit 0. It gets arbitrarily close, which is exactly what $\lim_{h \to 0}$ means. Now drag $h$ to a negative value so the second point sits to the left of the anchor. The secant approaches the same line and the readout approaches the same number. Left and right give one shared limit, which is what it means for $f'(a)$ to exist. Where a graph has a corner, the two sides would disagree and no single tangent slope would exist (Topic 2.4). NOTE (What the readout is proving): The label on the secant slope is average rate of change over the interval between $a$ and $a+h$. As you close the gap, that same number becomes the instantaneous rate of change at $a$. You are watching one quantity change its meaning in real time, which is the entire content of the limit definition. ### FAQ Q: What is the difference between a secant line and a tangent line? A: A secant line is the line through two chosen points on a curve, and its slope is the average rate of change between them. A tangent line meets the curve at the point of tangency, matching the curve's direction there (it may still cross the curve elsewhere), and its slope is the instantaneous rate of change at that point, which equals $f'(a)$. The tangent is the limiting position of the secant as the two points slide together. Q: Why can't I just plug in h = 0 to find the derivative? A: Substituting $h = 0$ turns the difference quotient into $\frac{f(a) - f(a)}{0} = \frac{0}{0}$, which is undefined, so the expression genuinely has no value at $h = 0$. The limit as $h \to 0$ sidesteps this by asking what value the quotient approaches for $h$ near 0 but not equal to 0. Q: Are the two limit definitions of the derivative the same? A: Yes. $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ and $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ are equal whenever the derivative exists. Substitute $x = a + h$: as $x \to a$, then $h \to 0$, and the two expressions match term for term. Use whichever form makes a given problem easier. ## Solid of Revolution Builder: Disk, Washer, and Shell URL: https://www.calclearn.org/interactives/solid-of-revolution-builder Last updated: 2026-07-22 Answer first: A solid of revolution is the shape made by spinning a flat region around an axis; its volume is a definite integral over thin slices: disks (pi r squared) for one curve on the axis, washers for a gap between two curves, shells for strips parallel to the axis. Drag the bounds and watch each integral assemble. A solid of revolution is what you get when you take a flat region in the plane and spin it a full turn around a straight line, the axis of revolution. The region sweeps out a 3D shape, and the question is almost always the same: what is its volume? The move that answers it is the one behind every integral. Slice the solid into thin pieces, write the volume of one piece, and add the pieces up with a definite integral (Topics 8.9 through 8.12). Where you take the slice decides the method. Cut perpendicular to the axis and each slice is a flat coin whose face is a full circle or a ring. Cut parallel to the axis and each slice is a thin cylindrical tube. Those two choices give the three formulas the builder assembles for you. | Method | Use it when | Slice area | Volume | | --- | --- | --- | --- | | Disk | one curve, the slice reaches the axis | $\pi\,[R(x)]^2$ | $\pi\int_a^b [R(x)]^2\,dx$ | | Washer | two curves leave a hole at the axis | $\pi\left([R]^2-[r]^2\right)$ | $\pi\int_a^b \left([R(x)]^2-[r(x)]^2\right)\,dx$ | | Shell | strip runs parallel to the axis | $2\pi\,x\,h(x)$ | $2\pi\int_a^b x\,h(x)\,dx$ | NOTE (What the AP exam actually requires): The CED builds volumes of revolution from the disk and washer methods only (Topics 8.9 through 8.12). The shell method is a standard technique many courses add for rotations about an axis parallel to the slices, but the CED does not test it. The builder includes shell so you can see the cleaner setup, and it names, on every region, the reason the other two methods do not fit. Choosing the method is the whole skill; the arithmetic is the easy part. ### How to read it Drag the two bound handles that sit on the x-axis (or press the arrow keys, holding Shift for fine steps) to change the interval the region sweeps over. - Only one method button lights up for each region. A single curve above the axis enables Disk, a second curve that carves a gap enables Washer, and a strip parallel to the axis enables Shell. Hover a greyed-out button to read exactly why that method does not fit this region. - Drag a bound off the region's own limits and the volume readout relabels itself from exact to numeric. At the full interval it prints the closed form, for example $8\pi \approx 25.13$ for $y=\sqrt{x}$; anywhere else it prints a composite-Simpson estimate for the sub-interval you dragged to. - On the washer region ($y=x$ over $y=x^2$) the middle panel carves a real hole down the axis, and the integrand is $x^2 - x^4$, the difference of the squared radii, not $(x-x^2)^2$. Squaring the gap is the single most common washer error. - Switch to the parabola-about-$y$ setup and the axis turns vertical: the solid redraws as nested cylindrical shells of radius $x$ and height $x^2$, and the entrance sweep replays each time you change regions. - The lower handle $a$ can never pass the upper handle $b$; a minimum gap is held so the interval stays valid no matter how far you drag. - Hold Shift while pressing an arrow key to nudge a bound by $0.01$ instead of $0.1$, so you can land the handle back onto the region's own limit and watch the numeric volume snap onto the exact value. The sine region's upper limit $\pi$ is the exception: it does not fall on the grid, so use the 'Reset bounds to the full region' button to return to the exact label there. When the region is bounded by a single curve and the axis itself, every slice taken perpendicular to the axis is a full disk. Its radius $R(x)$ is the distance from the axis out to the curve, and its face has area $\pi[R(x)]^2$. The first setup, $y=\sqrt{x}$ rotated about the x-axis, is the clean case: the radius is just $\sqrt{x}$, and squaring it collapses the integrand to $x$. $$V = \pi\int_0^4 \left(\sqrt{x}\right)^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi$$ The builder prints this as $V = 8\pi \approx 25.1327$ and stacks eleven of these disks along the axis so you can see the coin shape before the integral fuses them into one number. The fourth setup, $y=\sin x$ on $[0,\pi]$, is still a disk problem, but the slice area $\pi\sin^2 x$ needs a trig identity before it will integrate. $$V = \pi\int_0^{\pi} \sin^2 x\,dx = \pi\int_0^{\pi}\frac{1-\cos 2x}{2}\,dx = \frac{\pi}{2}\left[x-\frac{\sin 2x}{2}\right]_0^{\pi} = \frac{\pi^2}{2}$$ That lands on $\frac{\pi^2}{2} \approx 4.9348$, exactly the value the readout shows at the full bounds. Rewriting $\sin^2 x$ with the power-reduction identity is the same step you would reach for on a free-response volume question. When a second curve runs between the region and the axis, the slice is no longer solid; it is a ring, or washer. The outer radius $R(x)$ reaches the far curve and the inner radius $r(x)$ reaches the near one, so the ring's area is the big disk minus the hole. On the second setup, $y=x$ stays above $y=x^2$ across $[0,1]$, so the line gives the outer radius and the parabola gives the inner radius. $$V = \pi\int_0^1 \left(x^2 - \left(x^2\right)^2\right)\,dx = \pi\int_0^1 \left(x^2 - x^4\right)\,dx = \pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1 = \frac{2\pi}{15}$$ NOTE (Subtract the squares, not the gap): The integrand is $[R]^2-[r]^2 = x^2 - x^4$, never $(R-r)^2 = (x-x^2)^2$. Squaring the difference of the two radii is the most common washer mistake, and it gives a different, wrong volume. Square each radius on its own first, then subtract. The builder bakes $x^2 - x^4$ into the shown integral so you can check your own setup against it and confirm $\frac{2\pi}{15} \approx 0.4189$. Rotating the region under $y=x^2$ about the y-axis is the awkward case for disks: a slice perpendicular to the y-axis would force you to rewrite the curve as $x=\sqrt{y}$ and integrate in $y$. Slicing parallel to the axis avoids that. A thin vertical strip at position $x$ sweeps a cylindrical shell of radius $x$ and height $h(x)=x^2$; unrolled, the shell is a flat sheet of area $2\pi x \cdot h(x)$. $$V = 2\pi\int_0^2 x\cdot x^2\,dx = 2\pi\int_0^2 x^3\,dx = 2\pi\left[\frac{x^4}{4}\right]_0^2 = 8\pi$$ The shell method keeps $x$ as the variable and skips the inversion, which is why the builder marks it as the fitting method here. Remember, though, that the AP CED requires only disk and washer (Topics 8.9 through 8.12). On the exam this same region is handled as a washer integrated in $y$, with outer radius $2$ and inner radius $\sqrt{y}$: $$V = \pi\int_0^4 \left(2^2 - \left(\sqrt{y}\right)^2\right)\,dy = \pi\int_0^4 (4-y)\,dy = \pi\left[4y-\frac{y^2}{2}\right]_0^4 = 8\pi$$ Both routes land on $8\pi$, which is the sanity check worth internalizing: the volume of a solid does not depend on how you choose to slice it. Shell or washer, the number is the same; the methods only differ in how much algebra they cost. The bottom panel assembles the integral in stages. First the generic skeleton for the method, then the same integral with this region's curves and bounds filled in, then the exact closed form beside its decimal. Below that, three readouts track the current lower bound $a$, upper bound $b$, and the volume. At the region's own limits the volume is labeled exact and equals the closed form. Drag a bound anywhere else and the label flips to numeric: the builder switches to a composite-Simpson estimate of the volume over the sub-interval you chose, because no closed form is stored for an arbitrary interval. The closed form above keeps reporting the full-region value in its own limits, so the two are never confused. NOTE (Why the two numbers agree at the full bounds): Composite Simpson matches the exact value to the displayed precision on these polynomial and sine integrands, so when you drag the handles back onto the region's own limits the numeric estimate snaps onto the closed form. That agreement is the tool being honest about what it computes versus what it looks up: the decimal is always a real integral of the shown integrand, not a cached answer. ## Related Rates Scene: Sliding Ladder and Cone Tank URL: https://www.calclearn.org/interactives/related-rates-scene Last updated: 2026-07-22 Answer first: Related rates link two changing quantities with a geometric relation, then differentiate to connect their rates. The trick is order: differentiate while the quantities are still variables, and plug in the current numbers only at the end. A sliding ladder and a filling cone tank recompute the unknown rate as you drag. Every related rates problem starts with a relation, an equation that ties the changing quantities together and holds at every instant. Because it holds for all time, you can differentiate both sides with respect to $t$. The chain rule pulls a rate out of each quantity, and the single equation you get links all of those rates at once. This is implicit differentiation with time as the hidden variable (Topic 3.1, The Chain Rule and Topic 3.2, Implicit Differentiation, plus Topic 4.4, Introduction to Related Rates). $$x^2 + y^2 = 10^2 = 100$$ Differentiating with respect to $t$ brings a $\frac{dx}{dt}$ and a $\frac{dy}{dt}$ down through the chain rule, one term per changing length: $$2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0$$ NOTE (The order is the whole game): Notice that $x$ and $y$ are still symbols when you differentiate. If you substitute the current numbers first, say $x = 6$ and $y = 8$, then $x$ is a fixed constant, $\frac{dx}{dt}$ is zero, and the equation collapses to $0 = 0$ with nothing left to solve. Differentiate the relation while the quantities are variables, and plug in the state only at the end. The tool freezes the top four derivation rows for exactly this reason: the calculus is done before any number appears, and only the final substitution line moves when you drag. ### How to read it The scene puts the picture and its live derivation side by side. Drag the amber handle to slide the ladder's base along the floor, or switch to the cone tank and drag the water surface up and down, and the derivation's final line recomputes the unknown rate at that exact state. Everything above that line, the relation, its derivative, the given rate, and the solved formula, stays frozen, because all of the calculus is finished before any number is plugged in. A 10-foot ladder leans against a wall while its base slides away at $\frac{dx}{dt} = +2$ ft/s. The base distance $x$ and the wall height $y$ satisfy $x^2 + y^2 = 100$, so $y = \sqrt{100 - x^2}$. The question asks for $\frac{dy}{dt}$, the speed at which the top slides down. The scene walks the same five steps you use on any related rates free-response prompt: - Name the quantities and write the relation that ties them together. - Record the given rate and the rate you are solving for. - Differentiate the relation with respect to $t$, keeping every quantity a variable. - Substitute the known rate and the current values of the quantities. - Solve for the unknown rate and attach its units and sign. Differentiating and solving the rate as a formula in $x$ and $y$ gives: $$\frac{dy}{dt} = -\frac{x}{y}\,\frac{dx}{dt} = -\frac{2x}{y}$$ Only now does the current state go in. At $x = 6$ the height is exactly $y = \sqrt{100 - 36} = 8$, so: $$\frac{dy}{dt} = -\frac{2(6)}{8} = -\frac{3}{2} = -1.5 \text{ ft/s}$$ The sign is information, not decoration: $\frac{dy}{dt} < 0$ says the top is moving down. And the answer depends on where the ladder is, so sliding the base recomputes the last line while the four rows above it stay put. NOTE (Reading the ladder as you drag): $\frac{dy}{dt}$ is negative at every position (the top slides down) and grows in magnitude as the base moves out, from about $-0.2$ ft/s when the base is near the wall to about $-4.1$ ft/s at $x = 9$. At $x = 6$ the height is exactly $y = 8$, so the line reads an exact $\frac{dy}{dt} = -1.5$ ft/s with an equals sign; at most other positions $y$ is an irrational square root and the readout switches to the approximately-equal sign. The cone scene fills an inverted cone (height 10, top radius 5) at $\frac{dV}{dt} = +3$ cubic units per second and asks how fast the depth $h$ rises. A cone's volume is $V = \frac{1}{3}\pi r^2 h$, which carries two changing lengths, $r$ and $h$. Similar triangles pin them together: the water forms a smaller cone with the same proportions as the tank, so $\frac{r}{h} = \frac{5}{10}$, that is $r = \frac{h}{2}$. That constraint holds at every instant, so you may substitute it before differentiating. Doing so leaves the volume in a single variable, which is what makes the derivative clean: $$V = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h = \frac{\pi}{12}h^3$$ Differentiate with respect to $t$ and solve for the unknown rate: $$\frac{dV}{dt} = \frac{\pi}{4}h^2\,\frac{dh}{dt} \quad\Longrightarrow\quad \frac{dh}{dt} = \frac{12}{\pi h^2}$$ At $h = 2$ this gives $\frac{dh}{dt} = \frac{12}{4\pi} = \frac{3}{\pi} \approx 0.955$ units per second. Because $h^2$ sits in the denominator, the level climbs quickly when the cone is narrow near the tip and slows as the tank widens, even though $\frac{dV}{dt}$ never changes. Drag the surface up and watch $\frac{dh}{dt}$ fall. NOTE (Substituting the constraint is not the same as substituting the state): It is legitimate to plug $r = \frac{h}{2}$ in early because that relation is true for all time. What you must not do is plug in the instantaneous value $h = 2$ before differentiating, because that particular number is only true at one instant. The rule is about the changing state variable, not about relations that hold forever. NOTE (Reading the cone as you drag): $\frac{dh}{dt} = \frac{12}{\pi h^2}$ shrinks as $h$ rises: the surface climbs fast when the tank is narrow near the tip and slows as it widens, even though the fill rate $\frac{dV}{dt}$ stays fixed at $+3$. Every cone value carries the approximately-equal sign because the answer always contains $\pi$. The radius readout stays locked at $r = \frac{h}{2}$ for every depth, the similar-triangles constraint that lets you rewrite the volume in one variable before differentiating. Slide the ladder base toward the wall's far limit and $\frac{dy}{dt}$ races off the scale. The formula shows why: with $\frac{dx}{dt} = 2$, the rate is $\frac{dy}{dt} = \frac{-2x}{\sqrt{100 - x^2}}$, and as $x \to 10$ the height $y = \sqrt{100 - x^2}$ heads to $0$, so the denominator vanishes and the magnitude of $\frac{dy}{dt}$ grows without bound. $$\lim_{x \to 10^-} \frac{-2x}{\sqrt{100 - x^2}} = -\infty$$ The drag stops at $x = 9$ on purpose. A classic exam question asks for the rate at the instant the ladder lies flat, and the honest answer is that no finite rate exists there. Physically the top would have to move infinitely fast, the tell that the modeling assumption of a constant $\frac{dx}{dt}$ has broken down at the wall. Solving Related Rates Problems (Topic 4.5) rewards recognizing that the geometry, not the algebra, is what makes an instant impossible. - As you drag, only the bottom 'Substitute the current state' line changes; the relation, its derivative, the given rate, and the solved formula stay fixed, because all of the calculus is finished before any number is plugged in. - On the ladder, $\frac{dy}{dt}$ is negative at every position (the top slides down) and grows in magnitude as the base moves out, from about $-0.2$ ft/s when the base is near the wall to about $-4.1$ ft/s at $x = 9$. - At $x = 6$ the height is exactly $y = 8$, so the line reads an exact $\frac{dy}{dt} = -1.5$ ft/s with an equals sign; at most other positions $y$ is an irrational square root and the readout switches to the approximately-equal sign. - In the cone, $\frac{dh}{dt} = \frac{12}{\pi h^2}$ shrinks as $h$ rises: the surface climbs fast when the tank is narrow near the tip and slows as it widens, even though the fill rate $\frac{dV}{dt}$ stays fixed at $+3$. Every cone value carries the approximately-equal sign because the answer always contains $\pi$. - The radius readout stays locked at $r = \frac{h}{2}$ for every depth, the similar-triangles constraint that lets you rewrite the volume in one variable before differentiating. - The given rate is constant, but the unknown rate is a function of the current state: the same setup gives a different answer at every position, which is exactly why the substitution has to come last. --- # Comparisons: which one do I use ## Continuity vs Differentiability URL: https://www.calclearn.org/vs/continuity-vs-differentiability Last updated: 2026-07-28 Answer first: Differentiability is the stronger condition: every differentiable function is continuous, but plenty of continuous functions are not differentiable. Continuity means no break in the graph; differentiability additionally means no corner, cusp, or vertical tangent. - Continuity. Use when: You need the graph to have no hole, jump, or asymptote at the point, which is the hypothesis for the IVT and the EVT. - Differentiability. Use when: You need a single well-defined tangent slope, which is the hypothesis for the MVT and for any derivative computation. | | Continuity | Differentiability | | --- | --- | --- | | Requires | $\lim_{x \to a} f(x) = f(a)$ | The limit defining $f'(a)$ exists | | Broken by | Holes, jumps, asymptotes | All of those, plus corners, cusps, vertical tangents | | Implies the other | No | Yes, always | | Counterexample | None needed | $f(x) = \lvert x \rvert$ at $x = 0$ | The one-directional arrow is the whole comparison. If $f'(a)$ exists then $f$ must be continuous at $a$, because a jump would make the difference quotient blow up. Nothing forces the reverse: a graph can arrive at a point from both sides with no break and still have two different slopes there. The absolute value function is the permanent counterexample. At the origin it is continuous, since both one-sided limits and the function value all equal zero, but the slope is $-1$ on the left and $1$ on the right, so no single tangent slope exists. NOTE (For piecewise functions): Making a piecewise function differentiable at a seam takes two equations, not one. The pieces must meet, which is continuity, and their derivatives must match, which is differentiability. Solving only the first is the most common error. ### FAQ Q: Can a function be differentiable but not continuous? A: No. Differentiability at a point always implies continuity there, so a discontinuity rules out a derivative immediately. Q: Is a corner continuous? A: Yes. A corner is continuous but not differentiable, because the one-sided derivatives are different finite numbers. ## Removable vs Jump vs Infinite Discontinuity URL: https://www.calclearn.org/vs/removable-vs-jump-discontinuity Last updated: 2026-07-28 Answer first: A removable discontinuity is a hole where the limit exists but the value is missing or wrong. A jump has two different one-sided limits. An infinite discontinuity has at least one infinite one-sided limit. Only the removable kind can be repaired by redefining a single point. - Removable. Use when: Both one-sided limits agree but the function value is missing or different, so a factor cancelled. - Jump. Use when: Both one-sided limits exist and are finite but disagree, typical of a piecewise function whose pieces do not meet. - Infinite. Use when: At least one one-sided limit is infinite, so there is a vertical asymptote. | | Removable | Jump | Infinite | | --- | --- | --- | --- | | Two-sided limit | Exists | Does not exist | Does not exist | | Looks like | A hole | A step | An asymptote | | Can be repaired | Yes | No | No | | Typical source | A cancelling factor | A piecewise seam | A denominator zero that does not cancel | The deciding question is always about the two-sided limit. If it exists, the discontinuity is removable no matter what the function value is. If it fails because the sides disagree by a finite amount, it is a jump. If it fails because the function grows without bound, it is infinite. For a rational function the algebra tells you directly. Factor the numerator and the denominator: a factor that cancels leaves a hole, and a factor that survives in the denominator gives a vertical asymptote. NOTE (The mistake): Assuming every zero of the denominator is an asymptote. In $\frac{x^2 - 4}{x - 2}$ the denominator vanishes at 2, but the factor cancels, so the graph has a hole at $(2, 4)$ and no asymptote at all. ### FAQ Q: Which discontinuities still have a limit? A: Only removable ones. A jump and an infinite discontinuity both mean the two-sided limit does not exist. ## Average vs Instantaneous Rate of Change URL: https://www.calclearn.org/vs/average-vs-instantaneous-rate-of-change Last updated: 2026-07-28 Answer first: Average rate of change measures over an interval and equals the slope of the secant line joining the endpoints. Instantaneous rate of change measures at a single point and equals the slope of the tangent line, which is the derivative. - Average rate of change. Use when: The question names two times or two inputs, or asks how much something changed per unit over a stretch. - Instantaneous rate of change. Use when: The question names one moment, or uses the words at the instant, at time equals, or how fast right now. | | Average rate | Instantaneous rate | | --- | --- | --- | | Formula | $\frac{f(b) - f(a)}{b - a}$ | $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ | | Geometry | Secant slope | Tangent slope | | Needs calculus | No | Yes | | Inputs required | Two points | One point | The two are connected by a limit: shrink the interval and the average rate approaches the instantaneous rate. That is the entire construction of the derivative, and it is why the secant becomes the tangent as the second point slides in. The Mean Value Theorem is the other bridge. On a suitable interval, some interior point has an instantaneous rate exactly equal to the average rate across the whole interval. NOTE (Watch for a rate function): If you are already given a rate $r(t)$, then its average over an interval is $\frac{1}{b-a}\int_a^b r(t)\,dt$, which is the average value of a function and not the average rate of change of $r$. ### FAQ Q: Is average rate of change the same as average value? A: No. Average rate of change divides a change in output by a change in input. Average value integrates a function and divides by the interval length. ## Product Rule vs Chain Rule URL: https://www.calclearn.org/vs/product-rule-vs-chain-rule Last updated: 2026-07-28 Answer first: Use the product rule when two functions are multiplied side by side, and the chain rule when one function sits inside another. The test is to ask what you would compute last if you evaluated by hand: a multiplication means product rule, an outer function means chain rule. - Product rule. Use when: Two factors are multiplied and both contain the variable, as in $x^2 \sin x$. - Chain rule. Use when: One function is evaluated inside another, as in $\sin(x^2)$. | | Product rule | Chain rule | | --- | --- | --- | | Trigger | $f(x) \cdot g(x)$ | $f(g(x))$ | | Formula | $f g' + g f'$ | $f'(g(x)) \cdot g'(x)$ | | Example | $x^2 \sin x$ | $\sin(x^2)$ | | Result | $x^2\cos x + 2x\sin x$ | $2x\cos(x^2)$ | The two examples in the table differ only by where the parentheses sit, and that changes the answer completely. In $x^2 \sin x$ nothing is fed into anything, so the operations are independent. In $\sin(x^2)$ the squaring happens first and its output is handed to the sine. Real problems often need both. Differentiating $x^2 \sin(3x)$ starts with the product rule, and the chain rule then handles $\sin(3x)$ inside it. Deciding the outermost structure first keeps the work organized. NOTE (The mistake): Applying the chain rule to a product by differentiating the outer factor and multiplying by the derivative of the inner one. That is not a rule, and it gives the wrong answer for $x^2 \sin x$. ### FAQ Q: How do I know which rule is on the outside? A: Imagine plugging in a number and evaluating by hand. Whatever operation you would perform last is the outermost structure, and that decides the first rule you apply. ## Explicit vs Implicit Differentiation URL: https://www.calclearn.org/vs/explicit-vs-implicit-differentiation Last updated: 2026-07-28 Answer first: Differentiate explicitly when the equation is already solved for the output variable. Use implicit differentiation when the variables are mixed together and isolating one would be messy or impossible, as with a circle. - Explicit. Use when: The equation reads $y = $ something in $x$ alone, so you can differentiate directly. - Implicit. Use when: The equation mixes $x$ and $y$, as in $x^2 + y^2 = 25$ or $xy + \sin y = 4$. | | Explicit | Implicit | | --- | --- | --- | | Setup | $y = f(x)$ | An equation in $x$ and $y$ | | Method | Differentiate the right side | Differentiate both sides, then solve for $\frac{dy}{dx}$ | | Result depends on | $x$ only | Usually both $x$ and $y$ | | Evaluating a slope needs | An $x$ value | A full point | The mechanical difference is one factor. Every time you differentiate a term containing $y$, the chain rule contributes a $\frac{dy}{dx}$, because $y$ is itself a function of $x$. Differentiating $y^2$ gives $2y\frac{dy}{dx}$, not $2y$. Because the answer usually contains both variables, evaluating an implicit derivative requires a complete point rather than just an $x$ value. That is also why implicit answers can describe curves that fail the vertical line test. NOTE (When explicit is still better): If the equation solves cleanly, solving first is often less work. Implicit differentiation earns its keep when solving would introduce square roots and cases, as it does for a circle. ### FAQ Q: Why does dy/dx appear when I differentiate y? A: Because $y$ is a function of $x$, so the chain rule applies. Differentiating $y^3$ with respect to $x$ gives $3y^2 \frac{dy}{dx}$. ## Related Rates vs Optimization URL: https://www.calclearn.org/vs/related-rates-vs-optimization Last updated: 2026-07-28 Answer first: Related rates problems link two quantities changing over time and differentiate with respect to time. Optimization problems find a largest or smallest value and set a derivative equal to zero. Both start from a geometric relationship, but only optimization uses a constraint to eliminate a variable. - Related rates. Use when: Something is changing over time and the question gives you one rate and asks for another. - Optimization. Use when: The question asks for the largest, smallest, cheapest, or most efficient value, with nothing moving in time. | | Related rates | Optimization | | --- | --- | --- | | Differentiate with respect to | Time | The remaining variable | | Key step | Apply $\frac{d}{dt}$ to both sides | Set the derivative to zero | | Second equation | Not usually needed | The constraint, used to eliminate a variable | | Answer is | A rate, with units per time | A value or a location | The tell is time. If the problem says a ladder slides, water drains, or a balloon inflates, quantities depend on $t$ and you differentiate with respect to time. If nothing moves and the question asks for a best value, you are optimizing. Both share the same first step, which is writing a geometric relationship between the quantities. They diverge immediately after: related rates differentiates that relationship, while optimization uses a second constraint equation to reduce it to one variable first. NOTE (The shared mistake): Substituting numerical values too early. In related rates it turns a changing quantity into a constant whose derivative is zero. In optimization it collapses the function you were about to differentiate. ### FAQ Q: Do related rates problems need a constraint equation? A: Not in the optimization sense. They need one relationship between the quantities, which you then differentiate with respect to time. ## Local vs Absolute Extrema URL: https://www.calclearn.org/vs/local-vs-absolute-extrema Last updated: 2026-07-28 Answer first: A local extremum is the largest or smallest value compared only to nearby points. An absolute extremum is the largest or smallest across the entire interval. Local extrema occur only at critical points; absolute extrema occur at a critical point or at an endpoint. - Local extremum. Use when: The question asks where the function turns around, or asks for a relative maximum or minimum. - Absolute extremum. Use when: The question asks for the largest or smallest value on a specified closed interval. | | Local | Absolute | | --- | --- | --- | | Compared against | Nearby points only | Every point on the interval | | Can occur at an endpoint | No | Yes | | Found using | First or second derivative test | The candidates test | | How many can exist | Several | One maximum value, one minimum value | Every absolute extremum on a closed interval is either a local extremum or an endpoint value, which is exactly why the candidates test evaluates the function at every critical point and both endpoints and then compares. The Extreme Value Theorem is what guarantees absolute extrema exist at all, and it needs continuity on a closed interval. On an open interval a function can approach a value without ever attaining it. NOTE (The mistake): Comparing derivative values when hunting an absolute extremum. Critical points are located with $f'$, but the comparison that decides the answer uses $f$. ### FAQ Q: Can an endpoint be a local maximum? A: Under the usual convention, no. A local extremum needs points on both sides to compare against, so an endpoint can only hold an absolute extremum. ## First vs Second Derivative Test URL: https://www.calclearn.org/vs/first-vs-second-derivative-test Last updated: 2026-07-28 Answer first: The first derivative test classifies a critical point by whether the derivative changes sign there, and it never fails. The second derivative test checks the sign of the second derivative at the point, which is faster but says nothing when that value is zero. - First derivative test. Use when: You want a method that always reaches a conclusion, or the derivative does not exist at the point. - Second derivative test. Use when: The second derivative is easy to compute and you only need a quick classification. | | First derivative test | Second derivative test | | --- | --- | --- | | Uses | Sign of $f'$ on both sides | Sign of $f''$ at the point | | Maximum when | $f'$ goes positive to negative | $f''(c) < 0$ | | Can be inconclusive | No | Yes, when $f''(c) = 0$ | | Works where $f'$ is undefined | Yes | No | The second derivative test is really a statement about concavity. At a critical point on a concave down curve the graph must turn over, giving a maximum. When $f''(c) = 0$ the curve is not committing to a direction, so the test has nothing to say. Compare $x^4$, $-x^4$, and $x^3$ at the origin. All three have $f'(0) = f''(0) = 0$, yet they have a minimum, a maximum, and neither. Only the first derivative test separates them. NOTE (For justification credit): On free response, the credited justification for the first derivative test names the sign change of $f'$. Writing that $f'$ changes from positive to negative at $x = 2$ is what earns the point. ### FAQ Q: If the second derivative test is inconclusive, is there no extremum? A: No. Inconclusive means you need a different test. Fall back on the first derivative test, which always gives an answer. ## Critical Point vs Inflection Point URL: https://www.calclearn.org/vs/critical-point-vs-inflection-point Last updated: 2026-07-28 Answer first: A critical point is where the first derivative is zero or undefined, and it is where extrema can occur. An inflection point is where the second derivative changes sign, and it is where concavity flips. They answer different questions and often sit at different places. - Critical point. Use when: You are hunting maxima and minima, or describing where the function turns around. - Inflection point. Use when: You are describing where the graph changes its bend, or where a rate stops growing and starts shrinking. | | Critical point | Inflection point | | --- | --- | --- | | Derivative involved | $f'$ | $f''$ | | Condition | $f'$ is zero or undefined | $f''$ changes sign | | Tells you about | Extrema | Concavity | | Zero is enough | Yes, for candidacy | No, the sign must change | The two are independent. A function can have a critical point that is not an inflection point, as $x^2$ does at the origin, and an inflection point that is not critical, as $x^3 - 3x$ does at the origin where the slope is $-3$. Both come with the same trap in different forms. For a critical point, a zero derivative makes the point a candidate but not automatically an extremum. For an inflection point, a zero second derivative is not enough either, because the sign must actually change. NOTE (The classic counterexample): For $f(x) = x^4$ the second derivative is zero at the origin but stays positive on both sides, so the concavity never changes and there is no inflection point there. ### FAQ Q: Can a point be both critical and an inflection point? A: Yes. For $f(x) = x^3$ the origin has zero slope and a concavity change, so it is both, and it is not an extremum. ## Mean Value Theorem vs Intermediate Value Theorem URL: https://www.calclearn.org/vs/mvt-vs-ivt Last updated: 2026-07-28 Answer first: The Intermediate Value Theorem guarantees that a continuous function attains some output value, which is how you prove a root exists. The Mean Value Theorem guarantees that a differentiable function attains some slope, namely the average rate of change across the interval. - Intermediate Value Theorem. Use when: You need to prove an equation has a solution, or that the function equals some specific value somewhere. - Mean Value Theorem. Use when: You need to prove the derivative equals some value, often the average rate of change. | | IVT | MVT | | --- | --- | --- | | Guarantees a value of | $f$ | $f'$ | | Hypotheses | Continuous on $[a,b]$ | Continuous on $[a,b]$ and differentiable on $(a,b)$ | | Conclusion | $f(c) = N$ for some $c$ | $f'(c) = \frac{f(b)-f(a)}{b-a}$ | | Typical use | Proving a root exists | Proving a specific slope occurs | Reading the conclusion is the fastest way to tell them apart. If the statement to prove is about the function's output, reach for the IVT. If it is about a slope, a rate, or a derivative, reach for the MVT. The MVT asks for more because it promises more. It needs differentiability on the interior, which the IVT never requires, and that extra hypothesis must be stated explicitly for credit. NOTE (Both prove existence only): Neither theorem tells you where the point is or how many there are. Naming the theorem and verifying its hypotheses is the entire argument. ### FAQ Q: Which one proves an equation has a solution? A: The IVT. Rewrite the equation as $f(x) = 0$, show $f$ is continuous, and show the endpoint values have opposite signs. ## Mean Value Theorem vs Rolle's Theorem URL: https://www.calclearn.org/vs/mvt-vs-rolles-theorem Last updated: 2026-07-28 Answer first: Rolle's Theorem is the special case of the Mean Value Theorem where the two endpoint values are equal. When they match, the average rate of change is zero, so the guaranteed slope is zero and the tangent line is horizontal. - Mean Value Theorem. Use when: The endpoint values differ and you need a point whose slope equals the average rate of change. - Rolle's Theorem. Use when: The endpoint values are equal, often because both are roots, and you need a point with zero slope. | | MVT | Rolle's Theorem | | --- | --- | --- | | Extra hypothesis | None | $f(a) = f(b)$ | | Conclusion | $f'(c) = \frac{f(b)-f(a)}{b-a}$ | $f'(c) = 0$ | | Geometry | Tangent parallel to the secant | Horizontal tangent | | Typical use | Bounding a change in $f$ | Finding a critical point between two roots | They are the same theorem. Substituting $f(a) = f(b)$ into the MVT conclusion makes the fraction zero, which is exactly Rolle's statement, so anything Rolle's Theorem proves the MVT proves too. Rolle's Theorem earns its own name because of what it is used for: showing a critical point sits between two roots, which in turn bounds how many roots a function can have. NOTE (The shared trap): Both require differentiability on the open interval. The absolute value function on $[-1,1]$ has equal endpoint values but a corner at the origin, so neither theorem applies and no point has zero slope. ### FAQ Q: Do I ever need Rolle's Theorem if I know the MVT? A: Not strictly, since it is a special case. It is worth naming when the endpoint values are equal because the conclusion is cleaner. ## Speed vs Velocity URL: https://www.calclearn.org/vs/speed-vs-velocity Last updated: 2026-07-28 Answer first: Velocity is signed, so it tells you both how fast and in which direction. Speed is the absolute value of velocity and is never negative. Speed increases exactly when velocity and acceleration have the same sign, which is not the same as acceleration being positive. - Velocity. Use when: The question asks about direction, when the particle is at rest, or when it changes direction. - Speed. Use when: The question asks how fast, or whether the particle is speeding up or slowing down. | | Velocity | Speed | | --- | --- | --- | | Definition | $v(t) = s'(t)$ | $\lvert v(t) \rvert$ | | Can be negative | Yes | No | | Carries direction | Yes | No | | Integrates to | Displacement | Total distance travelled | The speeding-up rule is about agreement of signs, not the sign of acceleration alone. A particle with velocity $-4$ and acceleration $-2$ is speeding up, because it is moving in the negative direction faster and faster. The integration difference follows directly. Integrating velocity gives displacement, which can be small or zero after a round trip. Integrating speed gives total distance, which counts every unit travelled. ### FAQ Q: Does negative acceleration mean slowing down? A: Not necessarily. It means slowing down only when the velocity is positive. If velocity is also negative, the particle is speeding up. ## Displacement vs Total Distance Travelled URL: https://www.calclearn.org/vs/displacement-vs-distance-travelled Last updated: 2026-07-28 Answer first: Displacement is the net change in position, found by integrating velocity, and it can be zero even after a long trip. Total distance travelled integrates the absolute value of velocity, so movement in both directions adds up. - Displacement. Use when: The question asks for net change in position, or where the particle ends up relative to where it started. - Total distance. Use when: The question asks how far the particle travelled, or uses the phrase total distance. | | Displacement | Total distance | | --- | --- | --- | | Integral | $\int_a^b v(t)\,dt$ | $\int_a^b \lvert v(t) \rvert\,dt$ | | Can be zero | Yes | Only if the particle never moves | | Can be negative | Yes | No | | Needs sign analysis | No | Yes, split where $v$ changes sign | If a particle moves forward 5 units and back 3, the displacement is 2 and the distance travelled is 8. Whenever the velocity keeps one sign throughout the interval the two agree up to sign, which is why the distinction only bites when the particle turns around. To compute total distance by hand, find where the velocity changes sign, split the integral at those times, and add the absolute values of the pieces. On the calculator-active section you can integrate the absolute value directly. NOTE (Position from displacement): Displacement is a change, not a location. To get the final position, add it to the starting position: $s(b) = s(a) + \int_a^b v(t)\,dt$. ### FAQ Q: How do I know where velocity changes sign? A: Set $v(t) = 0$ and solve, then test the intervals between those times. Those are the moments the particle changes direction. ## Definite vs Indefinite Integral URL: https://www.calclearn.org/vs/definite-vs-indefinite-integral Last updated: 2026-07-28 Answer first: A definite integral has bounds and evaluates to a number. An indefinite integral has no bounds and evaluates to a family of functions plus a constant of integration. One is a value, the other is a function, and that single difference explains when the constant is needed. - Definite integral. Use when: You want an accumulated amount, an area, or a net change between two specific bounds. - Indefinite integral. Use when: You want the antiderivative itself, usually to solve a differential equation or to use later. | | Definite | Indefinite | | --- | --- | --- | | Notation | $\int_a^b f(x)\,dx$ | $\int f(x)\,dx$ | | Result | A number | A family of functions | | Needs $+ C$ | No | Yes | | Depends on the variable | No, it is a constant | Yes | The constant of integration disappears from a definite integral because it cancels in the subtraction: $\left(F(b) + C\right) - \left(F(a) + C\right)$ leaves $F(b) - F(a)$. That is why any antiderivative works for evaluating one. Because a definite integral is a number, the variable of integration is only a placeholder. The expressions $\int_a^b f(x)\,dx$ and $\int_a^b f(t)\,dt$ are the same number, which is not true of the indefinite versions. NOTE (The mistake): Leaving off the constant on an indefinite integral. Without it the answer names one antiderivative rather than all of them, and it is a routine deduction. ### FAQ Q: Can a definite integral be negative? A: Yes. It measures net signed accumulation, so region below the axis counts negatively. For true area, integrate the absolute value. ## FTC Part 1 vs Part 2 URL: https://www.calclearn.org/vs/ftc-part-1-vs-part-2 Last updated: 2026-07-28 Answer first: Part 1 differentiates an accumulation function and returns the integrand, which is the statement that differentiation undoes integration. Part 2 evaluates a definite integral as an antiderivative at the top bound minus the same antiderivative at the bottom. - FTC Part 1. Use when: You are asked to differentiate an integral whose upper bound contains the variable. - FTC Part 2. Use when: You are asked to compute the numerical value of a definite integral. | | Part 1 | Part 2 | | --- | --- | --- | | Statement | $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$ | $\int_a^b f(x)\,dx = F(b) - F(a)$ | | Operation | Differentiate | Evaluate | | Answer | A function | A number | | Watch for | A chain rule factor on the bound | Continuity across the interval | Part 1 is the reason accumulation functions can be analysed like any other function: the integrand is the derivative. Where $f$ is positive the accumulation increases, and where $f$ crosses from positive to negative the accumulation has a local maximum. Part 2 is what makes definite integrals computable without ever forming a Riemann sum. It needs the integrand to be continuous on the closed interval, so applying it blindly across a vertical asymptote produces nonsense. NOTE (The Part 1 trap): If the upper bound is a function rather than $x$ alone, the chain rule adds a factor. The derivative of $\int_0^{x^2} f(t)\,dt$ is $f(x^2) \cdot 2x$. ### FAQ Q: Which part do I use for an accumulation function question? A: Part 1, whenever you are differentiating. Use Part 2 when you need the numerical value at a specific input. ## U-Substitution vs Integration by Parts URL: https://www.calclearn.org/vs/u-substitution-vs-integration-by-parts Last updated: 2026-07-28 Answer first: Use substitution when the integrand contains a function and something close to its derivative, since substitution reverses the chain rule. Use integration by parts when the integrand is a product of two unlike types, such as a polynomial times an exponential, since parts reverses the product rule. - U-substitution. Use when: You can spot an inner function whose derivative is also present, up to a constant factor. - Integration by parts. Use when: The integrand multiplies two different kinds of function and no inner derivative is available. | | U-substitution | Integration by parts | | --- | --- | --- | | Reverses | The chain rule | The product rule | | Look for | $f(g(x)) \cdot g'(x)$ | A product of unlike types | | Formula | $\int f(u)\,du$ after substituting | $\int u\,dv = uv - \int v\,du$ | | Example | $\int 2x\cos(x^2)\,dx$ | $\int x e^x\,dx$ | Try substitution first, because it is faster and because failing costs nothing. If no candidate for $u$ leaves you with its own derivative sitting in the integrand, substitution stalls and parts becomes the next move. The two examples show the difference clearly. In $\int 2x\cos(x^2)\,dx$ the factor $2x$ is exactly the derivative of $x^2$, so substitution collapses it. In $\int x e^x\,dx$ neither factor is the derivative of the other, so parts is the way through. NOTE (Changing the bounds): For a definite integral, substitution requires you either to convert the bounds to the new variable or to convert back before evaluating. Mixing old bounds with a new variable is a routine and costly error. ### FAQ Q: How do I choose u for integration by parts? A: Pick $u$ so that differentiating it simplifies the problem, and $dv$ so that you can actually integrate it. A polynomial factor is usually a good $u$ because repeated differentiation eventually kills it. ## Left vs Right Riemann Sum URL: https://www.calclearn.org/vs/left-vs-right-riemann-sum Last updated: 2026-07-28 Answer first: Left and right Riemann sums differ only in whether each rectangle takes its height from the left or right endpoint of its subinterval. For an increasing function the left sum underestimates and the right sum overestimates, and for a decreasing function the roles swap. - Left Riemann sum. Use when: You are told to use left endpoints, or you want an underestimate for an increasing function. - Right Riemann sum. Use when: You are told to use right endpoints, or you want an overestimate for an increasing function. | | Left sum | Right sum | | --- | --- | --- | | Height from | Left endpoint of each subinterval | Right endpoint of each subinterval | | Skips which data point | The last one | The first one | | Increasing $f$ | Underestimate | Overestimate | | Decreasing $f$ | Overestimate | Underestimate | Monotonicity, not concavity, decides the direction of the error for these two. Concavity is what governs the trapezoidal and midpoint errors, so mixing the rules up is a common source of wrong justifications. Averaging the left and right sums gives exactly the trapezoidal rule, which is usually closer to the true value than either one alone. NOTE (The mistake): Counting data points as subintervals. A table with values at four times gives three subintervals, and if the spacing is uneven then $\Delta x$ is not constant and each rectangle needs its own width. ### FAQ Q: Which sum should I use if the function is not monotonic? A: Neither direction can be justified globally. Split the interval where the function changes direction, or use the trapezoidal rule and justify with concavity instead. ## Midpoint vs Trapezoidal Rule URL: https://www.calclearn.org/vs/midpoint-vs-trapezoidal-rule Last updated: 2026-07-28 Answer first: The trapezoidal rule joins consecutive points with straight chords, so on a concave up curve it overestimates. The midpoint rule takes each rectangle's height from the centre of its subinterval and errs in the opposite direction, underestimating on a concave up curve. - Trapezoidal rule. Use when: You have data at listed points and no values in between, which is the usual table situation. - Midpoint rule. Use when: You can evaluate the function anywhere, so you can sample the centre of each subinterval. | | Trapezoidal | Midpoint | | --- | --- | --- | | Shape used | Trapezoid | Rectangle at centre height | | Concave up | Overestimate | Underestimate | | Concave down | Underestimate | Overestimate | | Works from a table | Yes | Only if midpoints are listed | Both errors are governed by concavity, and they point opposite ways, which is why the true value is always trapped between them. On a concave up curve the chords sit above the graph while the midpoint rectangles cut under it. The trapezoidal rule also equals the average of the left and right Riemann sums, which is a fast way to compute it when you already have both. NOTE (Unequal widths): The compact trapezoidal formula assumes equal spacing. With data at uneven intervals, compute each trapezoid separately as the average of its two heights times its own width. ### FAQ Q: Which is more accurate? A: For a smooth function the midpoint rule is usually about twice as accurate as the trapezoidal rule, but it requires being able to evaluate the function at the midpoints. ## Area vs Net Signed Area URL: https://www.calclearn.org/vs/area-vs-net-signed-area Last updated: 2026-07-28 Answer first: A definite integral computes net signed area, so region below the horizontal axis counts as negative and can cancel region above it. True geometric area integrates the absolute value of the function, so every piece contributes positively. - Net signed area. Use when: The question asks for the value of the integral, a net change, or a displacement. - True area. Use when: The question asks for the area of a region, or uses the word area geometrically. | | Net signed area | True area | | --- | --- | --- | | Integral | $\int_a^b f(x)\,dx$ | $\int_a^b \lvert f(x) \rvert\,dx$ | | Below the axis counts as | Negative | Positive | | Can be zero | Yes | Only if $f$ is zero throughout | | Needs sign analysis | No | Yes, split at the zeros | The parallel with motion is exact. Net signed area is to true area what displacement is to total distance travelled, and both distinctions come from the same place: an integral adds signed quantities. To compute true area by hand, find where the function crosses the axis, split the integral at those points, and add the absolute values of the pieces. NOTE (Area between curves): When finding the area between two curves the integrand is top minus bottom, which is already non-negative, so no absolute value is needed as long as you know which curve is on top. ### FAQ Q: Does a negative integral mean I made a mistake? A: No. A negative value is a perfectly valid net signed area, and it is the right answer whenever more of the region sits below the axis. ## Average Value vs Average Rate of Change URL: https://www.calclearn.org/vs/average-value-vs-average-rate-of-change Last updated: 2026-07-28 Answer first: Average value integrates a function over an interval and divides by the interval length, answering what constant height would give the same area. Average rate of change divides the change in output by the change in input, answering how fast the function moved on average. - Average value. Use when: You are given a quantity and asked for its typical size over an interval. - Average rate of change. Use when: You are asked how much a quantity changed per unit of input across an interval. | | Average value | Average rate of change | | --- | --- | --- | | Formula | $\frac{1}{b-a}\int_a^b f(x)\,dx$ | $\frac{f(b) - f(a)}{b - a}$ | | Uses | An integral | Two function values | | Geometry | Height of an equal-area rectangle | Secant slope | | Related theorem | Mean Value Theorem for integrals | Mean Value Theorem | The trap is a rate function. If $r(t)$ is already a rate, then the average value of $r$ is $\frac{1}{b-a}\int_a^b r(t)\,dt$, and that is also the average rate of change of the accumulated quantity. Asking for the average rate of change of $r$ itself is a different question entirely. Both divide by $b - a$, which is what makes them averages rather than totals. Dropping that division is the single most common slip. ### FAQ Q: Is the average value ever attained by the function? A: For a continuous function on a closed interval, yes. That is the Mean Value Theorem for integrals. ## General vs Particular Solution URL: https://www.calclearn.org/vs/general-vs-particular-solution Last updated: 2026-07-28 Answer first: The general solution of a differential equation is the whole family of functions that satisfy it, written with an arbitrary constant. A particular solution is the single member of that family passing through a given initial condition. - General solution. Use when: No initial condition is given, or the question asks for the family of solutions. - Particular solution. Use when: An initial condition is given, such as a point the solution must pass through. | | General | Particular | | --- | --- | --- | | Contains | An arbitrary constant | A specific constant | | Describes | Infinitely many curves | One curve | | Needs an initial condition | No | Yes | | Matches a slope field | Every curve in it | One curve through the given point | The workflow is fixed: separate and integrate to get the general solution, substitute the initial condition to solve for the constant, then write the particular solution with that value in place. In a separable equation it is often cleaner to apply the initial condition right after integrating, before rearranging, because the constant is easier to isolate at that stage. NOTE (Solve for y): An answer left in implicit form is usually incomplete on the AP exam. Unless the question says otherwise, isolate the dependent variable before finishing. ### FAQ Q: How many initial conditions do I need? A: One for a first-order equation, which is all AP Calculus covers. Each integration introduces one constant that needs one condition to pin down. ## Exponential vs Logistic Growth URL: https://www.calclearn.org/vs/exponential-vs-logistic-growth Last updated: 2026-07-28 Answer first: Exponential growth has a rate proportional to the current amount and increases without any ceiling. Logistic growth multiplies that by a braking factor, so growth slows as the quantity nears a carrying capacity and is fastest at exactly half of it. - Exponential. Use when: The rate is proportional to the amount present, with no stated maximum. - Logistic. Use when: The problem names a carrying capacity, a limiting value, or a maximum sustainable population. | | Exponential | Logistic | | --- | --- | --- | | Equation | $\frac{dP}{dt} = kP$ | $\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)$ | | Long-run behaviour | Grows without bound | Approaches $L$ | | Fastest growth | At the largest $P$ | At $P = \frac{L}{2}$ | | Graph shape | J-shaped | S-shaped | The logistic factor $\left(1 - \frac{P}{L}\right)$ is the brake. When $P$ is small it is close to 1 and the curve looks exponential; as $P$ approaches $L$ it approaches zero and growth stalls. The inflection point at $P = \frac{L}{2}$ is where growth is fastest, and it is the fact most exam questions turn on. You can read it straight off the differential equation without solving it. NOTE (You rarely need the solved form): Carrying capacity, limiting behaviour, and fastest growth all come from the differential equation itself. Solving the logistic equation requires partial fractions and is seldom necessary. ### FAQ Q: What happens if the population starts above the carrying capacity? A: The braking factor turns negative, so the population decreases toward the carrying capacity instead of growing toward it. ## Slope Field vs Euler's Method URL: https://www.calclearn.org/vs/slope-field-vs-eulers-method Last updated: 2026-07-28 Answer first: A slope field is a picture: short segments showing the slope a differential equation assigns at each point, sketching every solution at once. Euler's method is a numerical procedure that starts from an initial condition and steps along tangent lines to produce actual approximate values. - Slope field. Use when: You need to sketch or identify solution behaviour, or match a field to an equation. - Euler's method. Use when: You need a numerical estimate of the solution at a specific input. | | Slope field | Euler's method | | --- | --- | --- | | Output | A picture | Numbers | | Needs an initial condition | No | Yes | | Shows | All solutions at once | One solution, approximately | | Error direction | Not applicable | Set by concavity | They are the same idea at different resolutions. Euler's method is what you get by actually walking along the slope field one small step at a time, using the equation to recompute the slope at each new point. Euler's error follows concavity: if the true solution is concave up, the tangent lines lie below it and the method underestimates. Smaller steps reduce the error but never remove it. NOTE (The Euler mistake): Reusing the initial slope for every step. The slope must be recomputed at each new point, which is the entire reason the method works. ### FAQ Q: Is Euler's method on the AB exam? A: No, Euler's method is BC only. Slope fields appear on both AB and BC. ## Disk vs Washer Method URL: https://www.calclearn.org/vs/disk-vs-washer-method Last updated: 2026-07-28 Answer first: Use the disk method when the region being rotated touches the axis of rotation, so each slice is a solid circle. Use the washer method when a gap separates the region from the axis, because that gap becomes a hole and each slice becomes a ring. - Disk method. Use when: The region is bounded by the axis of rotation itself, so no gap exists. - Washer method. Use when: The region is bounded away from the axis, or lies between two curves. | | Disk | Washer | | --- | --- | --- | | Slice shape | Solid circle | Ring | | Integrand | $\pi R^2$ | $\pi\left(R^2 - r^2\right)$ | | Region touches the axis | Yes | No | | Radii needed | One | Two | A disk is just a washer with inner radius zero, so there is really one method with a special case. Sketching the region and asking whether any gap separates it from the axis settles the choice in seconds. Both radii are distances from the axis, so rotating about a line other than an axis changes every expression. Rotating about $y = 4$ makes the outer radius $4 - f(x)$ rather than $f(x)$. NOTE (The mistake): Writing $(R - r)^2$ instead of $R^2 - r^2$. Those are different quantities, and only the second is the area of a ring. ### FAQ Q: How do I know which radius is outer? A: The outer radius reaches the curve farther from the axis of rotation. Sketch a representative slice and measure both distances from the axis. ## Washer vs Shell Method URL: https://www.calclearn.org/vs/washer-vs-shell-method Last updated: 2026-07-28 Answer first: Washers slice perpendicular to the axis of rotation and integrate in the axis variable. Shells slice parallel to it and integrate in the other variable. Choose whichever lets you avoid rewriting the equation in terms of a variable it is not already solved for. - Washer method. Use when: The curves are already expressed in the variable you would be integrating in for a perpendicular slice. - Shell method. Use when: A perpendicular slice would force you to solve the equation for the other variable. | | Washer | Shell | | --- | --- | --- | | Slice direction | Perpendicular to the axis | Parallel to the axis | | Integrand | $\pi\left(R^2 - r^2\right)$ | $2\pi r h$ | | Rotating about the $y$ axis | Integrate in $y$ | Integrate in $x$ | | Needs solving for the other variable | Often | Often not | Both give the same volume, so the choice is purely about which integral you can actually evaluate. Sketch a representative slice in each direction and see which one keeps the functions in the form you already have. The shell integrand is the circumference of the unrolled shell, $2\pi r$, times its height. Thinking of it as a flattened rectangle explains why there is no squaring and no subtraction. NOTE (Radius is still a distance): In both methods the radius is measured from the axis of rotation, so a shift in the axis changes the expression even though the region is unchanged. ### FAQ Q: Is the shell method on the AP exam? A: Shells are not required by the CED, but the method is often the fastest route and is fully valid on free response. ## Known Cross-Sections vs Solids of Revolution URL: https://www.calclearn.org/vs/cross-sections-vs-solids-of-revolution Last updated: 2026-07-28 Answer first: Both methods integrate cross-sectional area along an axis. A solid of revolution always has circular or ring-shaped slices because it is swept by rotation. A solid with known cross-sections can have squares, semicircles, or triangles instead, and the problem tells you which. - Solid of revolution. Use when: The problem says a region is rotated or revolved about a line. - Known cross-sections. Use when: The problem describes slices of a stated shape built on a base region. | | Revolution | Known cross-sections | | --- | --- | --- | | Slice shape | Circle or ring | Whatever the problem states | | Formula | $V = \pi\int R^2\,dx$ | $V = \int A(x)\,dx$ | | Key quantity | The radius | The base segment width | | Involves $\pi$ | Always | Only for circular shapes | Revolution is the special case where $A(x)$ happens to be $\pi R^2$. Once you see that, both problems reduce to the same job: write the area of a representative slice as a function of position, then integrate. For known cross-sections the base segment width does the work. Squares give $w^2$, equilateral triangles give $\frac{\sqrt{3}}{4}w^2$, and semicircles give $\frac{\pi w^2}{8}$ because the segment is the diameter. NOTE (The semicircle trap): When cross-sections are semicircles the base segment is the diameter, so the radius is half of it. Using the full width as the radius quadruples the answer. ### FAQ Q: Do I ever need pi for known cross-sections? A: Only when the stated shape is circular, such as a semicircle. Squares and triangles produce no pi at all. ## Parametric vs Polar Curves URL: https://www.calclearn.org/vs/parametric-vs-polar Last updated: 2026-07-28 Answer first: Parametric equations define the coordinates separately as functions of a parameter, usually time. Polar equations define distance from the origin as a function of angle. Polar is really a special case of parametric, which is why the slope formulas match. - Parametric. Use when: The problem describes motion, gives $x(t)$ and $y(t)$, or asks about a particle's path. - Polar. Use when: The curve is naturally described by distance and angle, such as a circle, cardioid, or rose. | | Parametric | Polar | | --- | --- | --- | | Definition | $x = x(t)$, $y = y(t)$ | $r = r(\theta)$ | | Slope | $\frac{dy/dt}{dx/dt}$ | Same formula with $\theta$ as the parameter | | Area | Not a standard AP topic | $\frac{1}{2}\int r^2\,d\theta$ | | Arc length | $\int \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$ | Same, after converting | Any polar curve becomes parametric by writing $x = r(\theta)\cos\theta$ and $y = r(\theta)\sin\theta$. That substitution is where the polar slope formula comes from, and it is worth remembering rather than memorizing a separate rule. Polar area is the one place the two genuinely differ. Sectors, not rectangles, are the natural slice, which produces the one half and the square in $\frac{1}{2}\int r^2\,d\theta$. NOTE (The mistake): Using $\int y\,dx$ for polar area. That formula assumes rectangular slices and gives the wrong answer for a region swept by an angle. ### FAQ Q: Can I always convert polar to parametric? A: Yes, using $x = r\cos\theta$ and $y = r\sin\theta$ with $\theta$ as the parameter. Converting is often the safest route for slope questions. ## Sequence vs Series URL: https://www.calclearn.org/vs/sequence-vs-series Last updated: 2026-07-28 Answer first: A sequence is an ordered list of numbers; a series is the sum of that list. A sequence converges when its terms approach a limit, while a series converges when its partial sums approach a limit, which is a much stronger requirement. - Sequence. Use when: You are asked what the terms approach, or whether the terms themselves settle down. - Series. Use when: You are asked whether a sum converges, or for the value of an infinite sum. | | Sequence | Series | | --- | --- | --- | | Object | A list | A sum | | Notation | $a_n$ | $\sum a_n$ | | Converges when | The terms approach a limit | The partial sums approach a limit | | Harmonic case | $\frac{1}{n} \to 0$, converges | $\sum \frac{1}{n}$ diverges | The harmonic case is the whole reason the distinction matters. The terms shrink to zero, so the sequence converges, yet the partial sums grow without bound and the series diverges. Terms going to zero is necessary for a series to converge but nowhere near sufficient. That asymmetry is exactly what the nth term test encodes: it can prove divergence when the terms do not approach zero, and it can never prove convergence. ### FAQ Q: If a sequence diverges, does its series diverge? A: Yes. If the terms do not approach zero the series cannot converge, which is the nth term test. ## Ratio Test vs Root Test URL: https://www.calclearn.org/vs/ratio-test-vs-root-test Last updated: 2026-07-28 Answer first: The ratio test compares consecutive terms and is the right choice when factorials or products of consecutive integers appear. The root test takes the nth root and is the right choice when the whole term is raised to the nth power. Both are inconclusive when the limit equals one. - Ratio test. Use when: The terms contain factorials, or each term is built from the previous one by multiplying. - Root test. Use when: The entire term is raised to the nth power, so taking an nth root cancels the exponent cleanly. | | Ratio test | Root test | | --- | --- | --- | | Limit taken | $\lim \left\lvert \frac{a_{n+1}}{a_n} \right\rvert$ | $\lim \sqrt[n]{\lvert a_n \rvert}$ | | Best for | Factorials | $n$th powers | | Converges when | $L < 1$ | $L < 1$ | | Inconclusive when | $L = 1$ | $L = 1$ | Both conclude absolute convergence when the limit is below one, so they handle series with mixed signs directly. Neither needs positive terms, which is why they are the default tools for power series. When one is inconclusive the other usually is too, because they agree whenever both limits exist. A limit of one means switching to a comparison, the integral test, or the alternating series test. NOTE (Where the ratio test really earns its place): Finding the radius of convergence of a power series. Applying it produces an inequality in the variable, and solving that inequality gives the radius. ### FAQ Q: Which should I try first? A: The ratio test, unless the whole term is visibly raised to the nth power. Factorials in particular collapse beautifully under the ratio test. ## Direct Comparison vs Limit Comparison Test URL: https://www.calclearn.org/vs/comparison-vs-limit-comparison-test Last updated: 2026-07-28 Answer first: Direct comparison requires you to prove a term by term inequality against a known series. Limit comparison only requires the two series to grow at comparable rates, judged by the limit of their ratio, which makes it easier to apply when the inequality is awkward. - Direct comparison. Use when: The inequality is obvious, such as dropping a positive term from a denominator. - Limit comparison. Use when: The terms clearly behave like a known series but the inequality points the wrong way or is hard to establish. | | Direct comparison | Limit comparison | | --- | --- | --- | | Requires | A proven inequality | A finite positive limit of the ratio | | Useful conclusions | Smaller than convergent, larger than divergent | Both series behave the same way | | Terms must be | Positive | Positive | | Fails when | The inequality goes the wrong way | The limit is 0 or infinite | Only two of the four possible direct comparisons say anything. Being smaller than a divergent series and larger than a convergent series are both uninformative, and that is where most errors with this test happen. Limit comparison sidesteps the problem. Choose the comparison series by keeping the dominant power on top and bottom: for $\sum \frac{3n^2 + 1}{n^4 - 2}$ that is $\frac{n^2}{n^4} = \frac{1}{n^2}$, a convergent p-series. NOTE (The comparison target): Almost always a p-series or a geometric series, because those are the only ones you can classify at sight. ### FAQ Q: What if the limit comparison limit is zero? A: The standard version is inconclusive. Pick a comparison series whose growth rate matches more closely and try again. ## Absolute vs Conditional Convergence URL: https://www.calclearn.org/vs/absolute-vs-conditional-convergence Last updated: 2026-07-28 Answer first: A series converges absolutely when the series of absolute values also converges. It converges conditionally when it converges as written but the absolute values diverge, meaning the convergence depends entirely on cancellation between positive and negative terms. - Absolute convergence. Use when: The series of absolute values converges, which also proves the original converges. - Conditional convergence. Use when: The alternating series test passes but the absolute values form a divergent series. | | Absolute | Conditional | | --- | --- | --- | | Original series | Converges | Converges | | Series of absolute values | Converges | Diverges | | Depends on cancellation | No | Yes | | Example | $\sum \frac{(-1)^n}{n^2}$ | $\sum \frac{(-1)^n}{n}$ | Deciding takes two tests, not one. Apply the alternating series test to the original, then test the absolute values separately. Convergent plus divergent means conditional; convergent plus convergent means absolute. The ratio and root tests are stated with absolute values, so whenever they conclude convergence they have already proved absolute convergence and no second test is needed. NOTE (Why absolute is stronger): Absolute convergence always implies convergence, so establishing it lets you use tests that require positive terms on a series with mixed signs. ### FAQ Q: Can a series of positive terms converge conditionally? A: No. With no negative terms the series and its absolute values are identical, so any convergence is absolute. ## Geometric Series vs p-Series URL: https://www.calclearn.org/vs/geometric-vs-p-series Last updated: 2026-07-28 Answer first: A geometric series has a constant ratio between consecutive terms, with the index in the exponent, and converges when the absolute ratio is below one. A p-series has the index in the base raised to a fixed power and converges when that power exceeds one. - Geometric series. Use when: Each term is a fixed multiple of the previous one, so the index sits in the exponent. - p-series. Use when: The terms are reciprocals of a power of the index, so the index sits in the base. | | Geometric | p-series | | --- | --- | --- | | Form | $\sum ar^n$ | $\sum \frac{1}{n^p}$ | | Index appears in | The exponent | The base | | Converges when | $\lvert r \rvert < 1$ | $p > 1$ | | Sum is known | Yes, $\frac{a}{1-r}$ | No | Both are classified at sight, which is why they are the comparison targets for every other test. Recognising which one you are looking at is a matter of asking whether $n$ is upstairs or downstairs. The geometric series is one of very few whose exact sum you can write down. A p-series tells you only whether it converges, not to what, with the single exception of specific known results outside the AP scope. NOTE (The geometric trap): In $\frac{a}{1-r}$ the value $a$ is the first term actually present. A sum starting at $n = 3$ has a different first term from one starting at $n = 0$. ### FAQ Q: Is the harmonic series geometric or a p-series? A: It is the p-series with $p = 1$, sitting exactly on the boundary, and it diverges. ## Taylor vs Maclaurin Series URL: https://www.calclearn.org/vs/taylor-vs-maclaurin-series Last updated: 2026-07-28 Answer first: A Maclaurin series is a Taylor series centred at zero. They are not different objects: the Maclaurin case is simply the most common centre, which is why the standard series for the exponential, sine, and cosine functions are all Maclaurin series. - Taylor series. Use when: The centre is any point, often chosen because the derivatives are easy there. - Maclaurin series. Use when: The centre is zero, which is the default unless the problem says otherwise. | | Taylor | Maclaurin | | --- | --- | --- | | Centre | Any $a$ | Zero | | Powers of | $(x - a)$ | $x$ | | Coefficient | $\frac{f^{(n)}(a)}{n!}$ | $\frac{f^{(n)}(0)}{n!}$ | | Relationship | The general case | The special case at $a = 0$ | Choosing a centre is a practical decision. To approximate $\ln(1.1)$ you would centre at 1, because the derivatives of $\ln x$ are clean there and zero is not even in the domain. In practice most series are built by substituting into a known Maclaurin series rather than differentiating repeatedly. The four worth memorizing are those for $e^x$, $\sin x$, $\cos x$, and $\frac{1}{1-x}$. ### FAQ Q: Does the centre change the interval of convergence? A: Yes. The interval is centred on the centre, so moving it moves the whole interval even when the radius is unchanged. ## Taylor Polynomial vs Taylor Series URL: https://www.calclearn.org/vs/taylor-polynomial-vs-taylor-series Last updated: 2026-07-28 Answer first: A Taylor polynomial is a finite truncation, so it approximates the function and carries an error you can bound. A Taylor series continues forever and, inside its interval of convergence, equals the function exactly. - Taylor polynomial. Use when: The question names a degree, asks for an approximation, or asks you to bound an error. - Taylor series. Use when: The question asks for the full series, a general term, or an interval of convergence. | | Polynomial | Series | | --- | --- | --- | | Number of terms | Finite | Infinite | | Relation to $f$ | Approximates | Equals, inside the interval | | Has an error | Yes | No, inside the interval | | Error bound | Lagrange, or first omitted term | Not applicable | The degree counts the highest power actually present, which can be lower than the number of terms you generated. For $\sin x$ only odd powers appear, so the second and third degree polynomials are the same expression. Error questions always concern the polynomial, never the series. Use the Lagrange error bound in general, or the first omitted term when the series alternates, which is much quicker. NOTE (The coefficient trap): The coefficient of $(x-a)^n$ is the $n$th derivative at the centre divided by $n!$. Dropping the factorial is the most common slip in building either object. ### FAQ Q: Does a Taylor series always equal the function? A: Inside its interval of convergence, for the functions in AP Calculus, yes. Outside the interval the series diverges and says nothing. ## Lagrange vs Alternating Series Error Bound URL: https://www.calclearn.org/vs/lagrange-vs-alternating-series-error-bound Last updated: 2026-07-28 Answer first: The alternating series error bound is the absolute value of the first omitted term, and it applies only when the series alternates and passes the alternating series test. The Lagrange error bound works for any Taylor polynomial but requires bounding the next derivative on the interval. - Alternating series bound. Use when: The series alternates and satisfies the alternating series test, which makes the bound a single term. - Lagrange error bound. Use when: The series does not alternate, or the question explicitly asks for the Lagrange bound. | | Alternating bound | Lagrange bound | | --- | --- | --- | | Formula | $\lvert a_{N+1} \rvert$ | $\frac{\max\lvert f^{(n+1)} \rvert}{(n+1)!}\lvert x-a \rvert^{n+1}$ | | Requires | An alternating series | A bound on the next derivative | | Effort | One term | Finding a maximum | | Tightness | Usually tight | Often loose | The alternating bound is simple because the partial sums overshoot and undershoot the true sum in turn, so the sum is always trapped between consecutive partial sums. That structure does not exist for a general series, which is why Lagrange needs more work. For sine and cosine the Lagrange bound simplifies a lot, since every derivative is bounded by one and the maximum is simply one. NOTE (The alternating trap): Use the first term you dropped, not the last one you kept. The bound is the size of the next term, the one not in your partial sum. ### FAQ Q: Is the Lagrange bound the actual error? A: No, it is an upper bound and is often much larger than the true error. Questions ask you to show the error is under some amount, which the bound does even when loose. ## Radius vs Interval of Convergence URL: https://www.calclearn.org/vs/radius-vs-interval-of-convergence Last updated: 2026-07-28 Answer first: The radius of convergence is a single number, the distance from the centre to where convergence stops, found with the ratio test. The interval of convergence is the full set of inputs where the series converges, which requires testing each endpoint separately. - Radius. Use when: The question asks only how far the series converges from its centre. - Interval. Use when: The question asks for all values of the variable where the series converges. | | Radius | Interval | | --- | --- | --- | | Answer type | A number | An interval with brackets | | Found with | The ratio test | The ratio test plus endpoint tests | | Endpoints | Not considered | Tested one at a time | | Ratio test at the endpoint | Not applicable | Always inconclusive there | The ratio test gives strict inequality, so it is silent exactly at the endpoints. That is not a flaw to work around; it is why endpoint testing is a separate step with a different test, usually the alternating series test on one side and a p-series comparison on the other. All four bracket combinations are possible, and the two endpoints very often behave differently, so each must be checked on its own. NOTE (The mistake): Reporting the radius and stopping when the question asked for the interval, or assuming both endpoints behave the same way. ### FAQ Q: Can the radius be zero or infinite? A: Yes. A radius of zero means the series converges only at its centre, and an infinite radius means it converges everywhere, as the series for $e^x$ does. ## Secant Line vs Tangent Line URL: https://www.calclearn.org/vs/secant-vs-tangent-line Last updated: 2026-07-28 Answer first: A secant line passes through two points on a curve and its slope is the average rate of change. A tangent line touches at a single point and its slope is the derivative there. Shrinking the gap between the two points turns the secant into the tangent. - Secant line. Use when: You have two points and want an average rate of change over the interval between them. - Tangent line. Use when: You have one point and want the instantaneous rate of change, or a linear approximation there. | | Secant | Tangent | | --- | --- | --- | | Points of contact | Two | One, locally | | Slope | $\frac{f(b)-f(a)}{b-a}$ | $f'(a)$ | | Needs calculus | No | Yes | | Related theorem | Mean Value Theorem | Linearization | The limit connecting them is the definition of the derivative. Writing the secant slope as $\frac{f(x+h) - f(x)}{h}$ and letting $h$ go to zero is exactly the difference quotient. The Mean Value Theorem ties them back together: on a suitable interval some tangent line is parallel to the secant line joining the endpoints. NOTE (A tangent can cross the curve): Touching at one point is a local description. A tangent line may intersect the curve again elsewhere, and at an inflection point it crosses right at the point of tangency. ### FAQ Q: Can a secant line and a tangent line be the same? A: For a straight line, yes, since every secant and tangent coincide with the line itself. ## Tangent Line vs Normal Line URL: https://www.calclearn.org/vs/tangent-vs-normal-line Last updated: 2026-07-28 Answer first: The tangent line at a point has slope equal to the derivative there. The normal line is perpendicular to it, so its slope is the negative reciprocal of the derivative. Both pass through the same point on the curve. - Tangent line. Use when: You need the direction the curve is heading, or a linear approximation near the point. - Normal line. Use when: The question explicitly asks for the normal, or for a line perpendicular to the curve. | | Tangent | Normal | | --- | --- | --- | | Slope | $f'(a)$ | $-\frac{1}{f'(a)}$ | | Relation | Parallel to the curve | Perpendicular to the curve | | If $f'(a) = 0$ | Horizontal | Vertical, $x = a$ | | Shared point | $(a, f(a))$ | $(a, f(a))$ | Everything carries over between them except the slope, which flips and changes sign. If the tangent slope is $\frac{2}{3}$, the normal slope is $-\frac{3}{2}$, and both lines use the same point in point-slope form. NOTE (The mistake): Using $f'(x)$ rather than $f'(a)$. The slope of a specific line is a number, obtained by evaluating the derivative at the point of tangency. ### FAQ Q: What is the normal line when the tangent is vertical? A: Horizontal. The normal line is $y = f(a)$, since perpendicular to vertical is horizontal. ## Vertical vs Horizontal Asymptote URL: https://www.calclearn.org/vs/vertical-vs-horizontal-asymptote Last updated: 2026-07-28 Answer first: A vertical asymptote occurs where a one-sided limit is infinite, and a graph can never cross one. A horizontal asymptote comes from a finite limit as the input grows without bound, describes long-run behaviour only, and can be crossed any number of times. - Vertical asymptote. Use when: You are looking at behaviour near a specific input where the function blows up. - Horizontal asymptote. Use when: You are looking at end behaviour as the input grows without bound. | | Vertical | Horizontal | | --- | --- | --- | | Comes from | $\lim_{x \to a} f(x) = \pm\infty$ | $\lim_{x \to \pm\infty} f(x) = L$ | | Equation | $x = a$ | $y = L$ | | Can be crossed | No | Yes | | How many possible | Any number | At most two | A graph cannot cross a vertical asymptote because there is no function value there at all. A horizontal asymptote is only a statement about the tail, so crossing it near the origin is perfectly ordinary. For rational functions, find vertical asymptotes at the denominator zeros that survive cancelling, and horizontal ones by comparing degrees: smaller on top gives $y = 0$, equal degrees give the ratio of leading coefficients. NOTE (The direction trap): As $x \to -\infty$, $\sqrt{x^2}$ equals $-x$, not $x$. Forgetting that flips the sign and is why some functions have two different horizontal asymptotes. ### FAQ Q: Why at most two horizontal asymptotes? A: Because there are only two directions to travel, so there are only two limits at infinity to take. ## Increasing vs Concave Up URL: https://www.calclearn.org/vs/increasing-vs-concave-up Last updated: 2026-07-28 Answer first: Increasing means the first derivative is positive, so the function is rising. Concave up means the second derivative is positive, so the slope itself is rising. The two are completely independent: a function can be increasing while concave down. - Increasing. Use when: You are describing whether the quantity is going up or down. - Concave up. Use when: You are describing whether the rate of change is going up or down. | | Increasing | Concave up | | --- | --- | --- | | Derivative involved | $f' > 0$ | $f'' > 0$ | | Describes | The function | The slope | | Plain language | Rising | Rising faster and faster | | Can happen together | Yes | Yes | All four combinations occur. A quantity growing but slowing down is increasing and concave down; a quantity falling ever faster is decreasing and concave down. Interpretation questions on free response are usually testing exactly this pairing. In context the second derivative is often the more interesting one. A company whose revenue is increasing but concave down is still growing, just less impressively each quarter. ### FAQ Q: If a function is concave up, must it be increasing? A: No. A parabola opening upward is concave up everywhere but decreases to the left of its vertex. ## Power Rule vs Exponential Rule URL: https://www.calclearn.org/vs/power-rule-vs-exponential-rule Last updated: 2026-07-28 Answer first: Use the power rule when the variable is the base and the exponent is a constant, as in x cubed. Use the exponential rule when the variable is the exponent and the base is a constant, as in two to the x. When the variable is in both places, neither rule works and you need logarithmic differentiation. - Power rule. Use when: The variable is the base and the exponent is a fixed number, including roots and reciprocals. - Exponential rule. Use when: The base is a fixed number and the variable sits in the exponent. | | Power rule | Exponential rule | | --- | --- | --- | | Form | $x^n$ | $a^x$ | | Derivative | $n x^{n-1}$ | $a^x \ln a$ | | Variable is | The base | The exponent | | Example | $x^5 \to 5x^4$ | $2^x \to 2^x \ln 2$ | The power rule is wider than it looks, because roots and reciprocals are powers in disguise: $\sqrt{x}$ is $x^{1/2}$ and $\frac{1}{x^3}$ is $x^{-3}$. Rewriting first turns both into one-step problems. The natural exponential is the clean special case. Since $\ln e = 1$, the derivative of $e^x$ is just $e^x$, which is why it shows up everywhere. NOTE (When the variable is in both places): For something like $x^x$ neither rule applies. Take the natural log of both sides and differentiate implicitly instead. ### FAQ Q: What is the derivative of x to the x? A: It is $x^x(\ln x + 1)$, found by logarithmic differentiation, because neither the power rule nor the exponential rule applies. --- # Glossary ## Absolute convergence URL: https://www.calclearn.org/glossary/absolute-convergence Last updated: 2026-07-28 Answer first: A series converges absolutely when the series formed by taking the absolute value of every term still converges. Absolute convergence is stronger than ordinary convergence and always implies it. The practical use is that it lets you apply tests requiring positive terms, such as the ratio and comparison tests, to a series with mixed signs. Establish absolute convergence and ordinary convergence follows for free. The ratio and root tests are stated in terms of absolute values, so when they conclude convergence they have proved absolute convergence. ## Acceleration URL: https://www.calclearn.org/glossary/acceleration Last updated: 2026-07-28 Answer first: Acceleration is the derivative of velocity with respect to time, and therefore the second derivative of position. Comparing its sign with the sign of velocity tells you whether the object is speeding up or slowing down. $$a(t) = v'(t) = s''(t)$$ The rule for speed is about agreement of signs, not the sign of acceleration alone. When velocity and acceleration share a sign the object speeds up; when the signs differ it slows down. NOTE (The mistake): Concluding that negative acceleration means slowing down. A particle with velocity $-4$ and acceleration $-2$ is speeding up, because it is moving in the negative direction faster and faster. ## Accumulation function URL: https://www.calclearn.org/glossary/accumulation-function Last updated: 2026-07-28 Answer first: An accumulation function is a definite integral whose upper bound is the variable, so it defines a new function measuring how much has accumulated from a fixed starting point. Its derivative is the integrand evaluated at the upper bound. Also called: Integral function $$g(x) = \int_a^x f(t)\,dt \implies g'(x) = f(x)$$ Analysing one is the same job as analysing any function, done with $f$ as the derivative. Where $f$ is positive $g$ increases, where $f$ crosses from positive to negative $g$ has a local maximum, and where $f$ increases $g$ is concave up. The dummy variable inside must differ from the bound. Writing $\int_a^x f(x)\,dx$ uses $x$ for two different roles and is why $t$ is the conventional choice inside. NOTE (With a chain rule): If the upper bound is a function rather than $x$ alone, the derivative picks up a factor: $\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x)) \cdot u'(x)$. ## Alternating series error bound URL: https://www.calclearn.org/glossary/alternating-series-error-bound Last updated: 2026-07-28 Answer first: For an alternating series that satisfies the alternating series test, the error in using a partial sum is less than the absolute value of the first omitted term. It is the simplest error bound in the course. $$\left|S - S_N\right| \le \left|a_{N+1}\right|$$ It works because the partial sums of an alternating series overshoot and undershoot the true sum in turn, so the sum is always trapped between consecutive partial sums. NOTE (The mistake): Using the last term you included rather than the first one you dropped. The bound is the size of the next term, the one not in your partial sum. ## Alternating series test URL: https://www.calclearn.org/glossary/alternating-series-test Last updated: 2026-07-28 Answer first: The alternating series test says that a series whose terms alternate in sign converges if the absolute values of its terms decrease and approach zero. Both conditions are needed, and together they are enough. Also called: Leibniz test Verify decrease explicitly rather than assuming it: either show $a_{n+1} \le a_n$ directly or show the corresponding function has a negative derivative. The test proves convergence only. To decide whether the convergence is absolute or conditional, apply another test to the series of absolute values. NOTE (The alternating harmonic series): It has decreasing terms going to zero, so it converges, while the harmonic series of absolute values diverges. That is the definition of conditional convergence. ## Antiderivative URL: https://www.calclearn.org/glossary/antiderivative Last updated: 2026-07-28 Answer first: An antiderivative of a function is any function whose derivative equals it. Because the derivative of a constant is zero, a function has infinitely many antiderivatives, all differing from one another by a constant. $$F'(x) = f(x) \iff F \text{ is an antiderivative of } f$$ The family of all antiderivatives is the indefinite integral, which is why the constant of integration appears there. On a connected interval, knowing one antiderivative determines all of them. Every antiderivative you find can be checked by differentiating it, which takes seconds and catches most sign and coefficient errors before they propagate. ## Arc length URL: https://www.calclearn.org/glossary/arc-length Last updated: 2026-07-28 Answer first: Arc length is the distance measured along a curve rather than straight across. It comes from adding up infinitesimal hypotenuses, which produces an integral of the square root of one plus the square of the derivative. $$L = \int_a^b \sqrt{1 + \left(f'(x)\right)^2}\,dx$$ Each tiny piece of curve is nearly a straight segment with horizontal leg $dx$ and vertical leg $dy$, so its length is $\sqrt{dx^2 + dy^2}$. Factoring out $dx$ gives the formula. In parametric form the same reasoning gives $L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$, which is also the integral of speed and therefore the total distance travelled. NOTE (Usually calculator work): Arc length integrands rarely have elementary antiderivatives, so these almost always appear on the calculator-active section. ## Average rate of change URL: https://www.calclearn.org/glossary/average-rate-of-change Last updated: 2026-07-28 Answer first: The average rate of change of a function over an interval is the change in output divided by the change in input, which is the slope of the secant line joining the endpoints. It needs no calculus, unlike the instantaneous rate of change. $$\frac{f(b) - f(a)}{b - a}$$ Average rate of change is a two-point calculation over an interval. Instantaneous rate of change is a one-point calculation and requires a limit. Confusing the two is the most frequent error on rate-interpretation questions. The Mean Value Theorem connects them: on a suitable interval, some interior point has an instantaneous rate exactly equal to the average rate across the whole interval. NOTE (For a rate function): If you are given a rate $r(t)$ and asked for its average value, that is $\frac{1}{b-a}\int_a^b r(t)\,dt$, which is a different quantity from the average rate of change of $r$. ## Average value of a function URL: https://www.calclearn.org/glossary/average-value-of-a-function Last updated: 2026-07-28 Answer first: The average value of a function on an interval is the definite integral over that interval divided by the length of the interval. It is the constant height a rectangle would need to have the same area as the region under the curve. Also called: Mean value of a function $$f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx$$ The rectangle picture is the useful one: you are flattening the region under the curve into a rectangle of the same width and equal area, and the average value is its height. NOTE (The mistake): Averaging the endpoint values, or forgetting to divide by $b - a$. Dividing by the interval length is what makes it an average rather than a total. ## Candidates test URL: https://www.calclearn.org/glossary/candidates-test Last updated: 2026-07-28 Answer first: The candidates test finds absolute extrema on a closed interval by evaluating the function at every critical point and at both endpoints, then comparing the resulting values. The largest is the absolute maximum and the smallest is the absolute minimum. Also called: Closed interval method It works because the Extreme Value Theorem guarantees the extrema exist and restricts where they can be. With only finitely many candidates, comparing function values settles the question completely. NOTE (The mistake): Comparing derivative values instead of function values. The candidates are located using $f'$, but the comparison that decides the answer uses $f$. ## Comparison test URL: https://www.calclearn.org/glossary/comparison-test Last updated: 2026-07-28 Answer first: The direct comparison test settles a series of positive terms by bounding it against a series whose behaviour is known. If it is term by term smaller than a convergent series it converges, and if it is larger than a divergent series it diverges. Also called: Direct comparison test Only two of the four possible comparisons say anything. Being smaller than a divergent series and larger than a convergent series are both uninformative, which is where most errors happen. The usual comparison targets are p-series and geometric series, since those are the ones you can classify at sight. NOTE (When the inequality goes the wrong way): If you cannot get the inequality you need, switch to the limit comparison test, which only requires the terms to behave alike rather than be ordered. ## Composite function URL: https://www.calclearn.org/glossary/composite-function Last updated: 2026-07-28 Answer first: A composite function is one function evaluated inside another, so the output of the inner function becomes the input of the outer one. Spotting a composition is what tells you to use the chain rule. Also called: Function composition $$(f \circ g)(x) = f(g(x))$$ The practical test is to ask what you would compute last if you evaluated the expression by hand. In $\sin(x^2)$ you would square first and take the sine last, so the outer function is sine and the inner function is $x^2$. NOTE (The mistake): Reading a product as a composition. In $x^2 \sin x$ nothing is fed into anything, so that is the product rule, not the chain rule. ## Concavity URL: https://www.calclearn.org/glossary/concavity Last updated: 2026-07-28 Answer first: Concavity describes which way a graph bends. A graph is concave up where the second derivative is positive and the slope is increasing, and concave down where the second derivative is negative and the slope is decreasing. Also called: Concave up, Concave down The clearest way to hold it is in terms of the tangent line: where a curve is concave up it lies above its tangent lines, and where it is concave down it lies below them. That fact also tells you whether a linear approximation overestimates or underestimates. NOTE (The mistake): Tying concavity to whether the function is increasing. They are independent: a function can be increasing while concave down, which is what a decelerating but still rising quantity looks like. ## Conditional convergence URL: https://www.calclearn.org/glossary/conditional-convergence Last updated: 2026-07-28 Answer first: A series converges conditionally when it converges as written but the series of its absolute values diverges. The cancellation between positive and negative terms is what makes it work, so the convergence depends on the order of the terms. The deciding procedure is two tests, not one: apply the alternating series test to the original series, then test the absolute values separately. Convergent plus divergent means conditional; convergent plus convergent means absolute. The alternating harmonic series is the standard example, converging to $\ln 2$ while its absolute values form the divergent harmonic series. ## Constant multiple rule URL: https://www.calclearn.org/glossary/constant-multiple-rule Last updated: 2026-07-28 Answer first: The constant multiple rule says a constant factor passes straight through a derivative, and the sum rule says the derivative of a sum is the sum of the derivatives. Together they make differentiation linear. Also called: Sum rule $$\frac{d}{dx}\left[c f(x) + g(x)\right] = c f'(x) + g'(x)$$ Linearity is why long polynomials are differentiated term by term with no extra machinery. It also carries over to integration, where constants factor out and sums split apart in exactly the same way. NOTE (Sums only): Linearity applies to sums and constant multiples, never to products or quotients of two functions of the variable. Those need their own rules. ## Constant of integration URL: https://www.calclearn.org/glossary/constant-of-integration Last updated: 2026-07-28 Answer first: The constant of integration is the arbitrary constant added to an indefinite integral. It is required because differentiating any constant gives zero, so a function has infinitely many antiderivatives that differ only by a constant. Also called: Plus C It stops being arbitrary the moment you have an initial condition. In a differential equation, substituting a known point determines the constant and turns the general solution into a particular one. NOTE (Not needed for definite integrals): In a definite integral the constant cancels when you subtract the two endpoint values, which is why it never appears in the final answer there. ## Continuity URL: https://www.calclearn.org/glossary/continuity Last updated: 2026-07-28 Answer first: A function is continuous at a point when three things hold: the function is defined there, the limit exists there, and the limit equals the function value. Informally, you can draw the graph through that point without lifting your pencil. Also called: Continuous function $$\lim_{x \to a} f(x) = f(a)$$ The compact statement $\lim_{x \to a} f(x) = f(a)$ quietly contains all three conditions: writing $f(a)$ requires it to exist, writing the limit requires it to exist, and the equals sign requires them to match. Exam questions often ask you to verify each part separately. A function is continuous on an interval when it is continuous at every point of that interval, using one-sided limits at the endpoints. Continuity on a closed interval is the hypothesis that unlocks the Intermediate Value Theorem and the Extreme Value Theorem. NOTE (The mistake): Treating continuity and differentiability as the same thing. Differentiability implies continuity, but never the reverse: $f(x) = |x|$ is continuous at 0 and not differentiable there. ## Convergence URL: https://www.calclearn.org/glossary/convergence Last updated: 2026-07-28 Answer first: A series converges when its sequence of partial sums approaches a finite limit, and diverges when it does not. Divergence covers both partial sums that grow without bound and ones that oscillate without settling. Also called: Divergence, Convergent, Divergent Convergence is a property of the tail, not the front. Changing or removing finitely many terms cannot change whether a series converges, though it does change the sum. NOTE (Say which): On free response, a bare claim of convergence earns nothing. Name the test, verify its conditions, and state the conclusion. ## Corner URL: https://www.calclearn.org/glossary/corner Last updated: 2026-07-28 Answer first: A corner is a point where a graph changes direction abruptly, with left and right derivatives that are different finite numbers. The function is continuous there but not differentiable, because no single tangent slope exists. The absolute value function has a corner at the origin: the slope is $-1$ approaching from the left and $1$ from the right. Both one-sided derivatives exist and are finite, which is exactly what distinguishes a corner from a cusp. Corners are common at the seams of piecewise functions and anywhere an absolute value appears, so check those points before claiming a function is differentiable everywhere. ## Critical point URL: https://www.calclearn.org/glossary/critical-point Last updated: 2026-07-28 Answer first: A critical point of a function is an interior point of its domain where the derivative is either zero or undefined. Every local maximum and minimum occurs at a critical point, though not every critical point is an extremum. Also called: Critical number, Critical value Both halves of the definition matter. Setting $f'(x) = 0$ finds the smooth turning points, but points where $f'$ fails to exist, such as the corner of $|x|$ at the origin, are critical too and are the ones most often missed. Critical points are candidates, not conclusions. The function $f(x) = x^3$ has $f'(0) = 0$ yet no extremum there, because the derivative does not change sign. NOTE (The mistake): Counting a point where the denominator of $f'$ vanishes but which is outside the domain of $f$. A critical point must be in the domain of the original function. ## Cross-section URL: https://www.calclearn.org/glossary/cross-section Last updated: 2026-07-28 Answer first: A cross-section is the two-dimensional shape you get by slicing a solid perpendicular to an axis. If you know the area of each cross-section as a function of position, integrating that area function along the axis gives the volume. Also called: Known cross-sections $$V = \int_a^b A(x)\,dx$$ The whole problem is writing $A(x)$. For squares built on a base of width $w$ the area is $w^2$; for semicircles of diameter $w$ it is $\frac{\pi w^2}{8}$; for equilateral triangles of side $w$ it is $\frac{\sqrt{3}}{4}w^2$. NOTE (Diameter or radius): When cross-sections are semicircles the base segment is the diameter, so the radius is half of it. Substituting the full width as the radius is the classic error and quadruples the answer. ## Cusp URL: https://www.calclearn.org/glossary/cusp Last updated: 2026-07-28 Answer first: A cusp is a sharp point on a graph where the one-sided slopes grow without bound in opposite directions. The function stays continuous there but is not differentiable, because the tangent slope is not a finite number. The function $f(x) = x^{2/3}$ has a cusp at the origin. Its derivative $\frac{2}{3}x^{-1/3}$ goes to $-\infty$ from the left and $+\infty$ from the right, so the graph comes to a point rather than a wedge. NOTE (Cusp versus corner): At a corner the one-sided slopes are finite and merely unequal. At a cusp they are infinite and opposite in sign. Both block differentiability. ## Definite integral URL: https://www.calclearn.org/glossary/definite-integral Last updated: 2026-07-28 Answer first: A definite integral is a number defined as the limit of Riemann sums as the subintervals shrink to zero width. It represents the net signed accumulation of a quantity between two bounds, with area below the axis counting as negative. $$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x$$ The word signed is doing real work. Region below the horizontal axis contributes negatively, so a definite integral can be zero or negative even though it is often described as area. For genuine area you integrate the absolute value. The Fundamental Theorem of Calculus is what makes these computable without ever forming a Riemann sum, by connecting the definite integral to any antiderivative of the integrand. NOTE (It is a number): The variable of integration is a placeholder, so $\int_a^b f(x)\,dx$ and $\int_a^b f(t)\,dt$ are the same number. Nothing in the answer depends on the letter used. ## Derivative URL: https://www.calclearn.org/glossary/derivative Last updated: 2026-07-28 Answer first: The derivative of a function at a point is its instantaneous rate of change there, defined as the limit of the difference quotient as the gap shrinks to zero. Geometrically it is the slope of the tangent line at that point. Also called: First derivative $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ Three descriptions of the derivative appear on the exam and all mean the same thing: the slope of the tangent line, the instantaneous rate of change, and the limit of average rates of change. Being able to move between them is most of what free-response questions test. The derivative is itself a function. Evaluating it at a specific input gives a number, the slope at that one point, which is why $f'(x)$ and $f'(2)$ mean different kinds of object. NOTE (Units matter): If $f$ measures gallons and $x$ measures minutes, then $f'$ is in gallons per minute. Interpretation questions award points for the units and the meaning in context, not the number alone. ## Difference quotient URL: https://www.calclearn.org/glossary/difference-quotient Last updated: 2026-07-28 Answer first: The difference quotient is the slope of the secant line joining two points on a curve, written as the change in output over the change in input. Taking its limit as the gap shrinks to zero produces the derivative. Also called: Newton quotient $$\frac{f(x+h) - f(x)}{h}$$ The expression is exactly rise over run for the two points $(x, f(x))$ and $(x+h, f(x+h))$. Every piece of derivative notation traces back to it, which is why the AP exam still asks you to compute derivatives from the definition. The alternate form $\frac{f(x) - f(a)}{x - a}$ describes the same slope with the second point named directly rather than as an offset. Its limit as $x \to a$ gives $f'(a)$. NOTE (The mistake): Writing $f(x+h)$ as $f(x) + h$. You must substitute $x+h$ into the whole rule, so for $f(x) = x^2$ you get $(x+h)^2 = x^2 + 2xh + h^2$. ## Differentiability URL: https://www.calclearn.org/glossary/differentiability Last updated: 2026-07-28 Answer first: A function is differentiable at a point when the limit defining the derivative exists there, which requires the graph to be locally smooth with a single well-defined tangent slope. Differentiability always implies continuity, but continuity does not imply differentiability. Also called: Differentiable Differentiability fails in four recognizable ways: a corner, where the one-sided slopes disagree; a cusp, where they run off to opposite infinities; a vertical tangent, where the slope is infinite; and any discontinuity, which rules out a derivative immediately. The implication runs one way only. Because $|x|$ is continuous at 0 but has slopes of $-1$ and $1$ on either side, it is a permanent counterexample to the converse. NOTE (For piecewise functions): Making a piecewise function differentiable at a seam takes two equations, not one: the pieces must meet, and their derivatives must match there. ## Differential URL: https://www.calclearn.org/glossary/differential Last updated: 2026-07-28 Answer first: The differential dy is the change in output along the tangent line corresponding to a small change dx in the input. It approximates the true change in the function and is the notation that makes substitution and separable equations work. Also called: dx, dy $$dy = f'(x)\,dx$$ The distinction worth holding onto is between $\Delta y$, the exact change in the function, and $dy$, the change along the tangent line. They agree closely when $dx$ is small, which is exactly the claim linearization makes. Differentials are also why u-substitution is written the way it is. Setting $u = g(x)$ gives $du = g'(x)\,dx$, which is what lets you replace one differential with another inside an integral. ## Differential equation URL: https://www.calclearn.org/glossary/differential-equation Last updated: 2026-07-28 Answer first: A differential equation is an equation that relates an unknown function to its own derivatives. Its solution is a function rather than a number, and there is usually a whole family of them until an initial condition picks one out. The equation $\frac{dy}{dx} = 2x$ asks which function has derivative $2x$, and the answer is the family $y = x^2 + C$. Verifying a proposed solution is a matter of differentiating it and checking the equation holds. AP Calculus covers separable equations, which are the ones you can rearrange so each variable sits with its own differential, plus slope fields and Euler's method for the ones you cannot solve exactly. ## Direct substitution URL: https://www.calclearn.org/glossary/direct-substitution Last updated: 2026-07-28 Answer first: Direct substitution evaluates a limit by putting the value straight into the function. It is valid exactly when the function is continuous at that point, which makes it the first thing to try on any limit. If substitution gives a number, that number is the limit and you are finished. If it gives an indeterminate form, the attempt has told you which technique to reach for next rather than wasted your time. A result like $\frac{5}{0}$ is not indeterminate: it means the limit is infinite, so check the one-sided behaviour to determine the sign. ## Disk method URL: https://www.calclearn.org/glossary/disk-method Last updated: 2026-07-28 Answer first: The disk method computes the volume of a solid of revolution by slicing it into circular disks perpendicular to the axis of rotation and integrating the area of each. It applies when the region being rotated touches the axis, leaving no hole. $$V = \pi \int_a^b \left(R(x)\right)^2 dx$$ Each slice is a circle of radius equal to the distance from the axis to the curve, so its area is $\pi R^2$. Integrating those areas along the axis accumulates the volume. NOTE (The mistake): Squaring only part of the radius. When rotating about a line such as $y = 2$, the radius is $2 - f(x)$ and the entire expression must be squared. ## Displacement URL: https://www.calclearn.org/glossary/displacement Last updated: 2026-07-28 Answer first: Displacement is the net change in position over a time interval, computed by integrating velocity. Total distance travelled integrates the absolute value of velocity instead, so it counts backtracking as additional distance. Also called: Total distance travelled $$\text{displacement} = \int_a^b v(t)\,dt, \qquad \text{distance} = \int_a^b |v(t)|\,dt$$ To compute total distance by hand, find where velocity changes sign, split the integral at those times, and add the absolute values of the pieces. On a calculator you can integrate the absolute value directly. NOTE (The mistake): Reporting displacement when the question asks for distance travelled. If the velocity changes sign anywhere on the interval, the two answers differ. ## Domain URL: https://www.calclearn.org/glossary/domain Last updated: 2026-07-28 Answer first: The domain of a function is the set of inputs for which it produces a value. Domain restrictions determine where a function can be continuous or differentiable, and they are what make some answers to calculus problems invalid. The usual restrictions are denominators that cannot be zero, square roots that need non-negative arguments, and logarithms that need strictly positive arguments. Each one creates a place where continuity must be checked. NOTE (Check the answer against it): Optimization and related rates problems often produce algebraically valid solutions that fall outside the domain, such as a negative length. Discarding those is part of the work. ## Epsilon-delta definition URL: https://www.calclearn.org/glossary/epsilon-delta-definition Last updated: 2026-07-28 Answer first: The epsilon-delta definition states that a limit equals L if, for every tolerance around L, there is a distance around the input point that keeps all outputs within that tolerance. It is the precise version of the informal idea of approaching a value. Also called: Formal definition of a limit $$\forall \varepsilon > 0, \ \exists \delta > 0: \ 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon$$ Read it as a challenge and a response: someone names a tolerance, and you must produce a closeness that guarantees it. The limit exists when you can always answer, no matter how small the tolerance. The condition $0 < |x - a|$ deliberately excludes the point itself, which is the formal statement of the idea that a limit ignores the function's value at the point. NOTE (On the AP exam): You are expected to understand what the definition says, not to construct epsilon-delta proofs. The computational work is done with limit laws and algebra. ## Euler's method URL: https://www.calclearn.org/glossary/eulers-method Last updated: 2026-07-28 Answer first: Euler's method approximates the solution of a differential equation by starting at a known point and repeatedly stepping along the tangent line for a small increment. Each step uses the differential equation to compute the slope at the current point. $$y_{n+1} = y_n + f(x_n, y_n)\,\Delta x$$ It is linearization applied over and over. Each step is exactly a tangent line approximation, and the new endpoint becomes the starting point for the next one. The error direction follows concavity: if the true solution is concave up, the tangent lines lie below it and Euler's method underestimates. Smaller steps reduce the error but never eliminate it. NOTE (The mistake): Reusing the initial slope for every step. The slope must be recomputed at each new point, which is the entire reason the method works. ## Exponential growth URL: https://www.calclearn.org/glossary/exponential-growth Last updated: 2026-07-28 Answer first: Exponential growth describes a quantity whose rate of change is proportional to its current amount. The differential equation says the derivative equals a constant times the quantity, and its solution is a constant multiple of e raised to that constant times time. Also called: Exponential decay $$\frac{dy}{dt} = ky \implies y = Ce^{kt}$$ The sign of $k$ decides everything: positive means growth, negative means decay. The constant $C$ is the initial amount, because setting $t = 0$ leaves $y = C$. This model appears as population growth, radioactive decay, and continuously compounded interest. Recognizing the phrase proportional to the amount present is what signals it. NOTE (Unbounded): Pure exponential growth has no ceiling, which is why realistic population models switch to logistic growth once resources become limited. ## Extreme Value Theorem URL: https://www.calclearn.org/glossary/extreme-value-theorem Last updated: 2026-07-28 Answer first: The Extreme Value Theorem says that a function continuous on a closed, bounded interval must attain both an absolute maximum and an absolute minimum somewhere on that interval. It guarantees the extrema exist without saying where they are. Also called: EVT Both hypotheses carry weight. On the open interval $(0, 1)$ the function $f(x) = x$ attains neither extreme, and a discontinuous function on a closed interval can also miss them. Drop either condition and the guarantee is gone. Because the EVT tells you the extrema exist and must sit at a critical point or an endpoint, it is what licenses the candidates test: evaluate the function at every critical point and both endpoints, then compare. ## Extremum URL: https://www.calclearn.org/glossary/extremum Last updated: 2026-07-28 Answer first: An extremum is a maximum or minimum value of a function. A local extremum is largest or smallest compared only to nearby points, while an absolute extremum is largest or smallest across the entire domain or interval under consideration. Also called: Maximum, Minimum, Extreme value Local extrema occur only at critical points. Absolute extrema on a closed interval occur either at a critical point or at an endpoint, which is why the candidates test checks both. The distinction between the location and the value is worth guarding. The extremum is the output value; the point where it happens is the $x$ coordinate. Questions ask for one or the other and rarely both. NOTE (Endpoints): An endpoint can hold an absolute extremum but never a local one under the usual convention, because a local extremum needs points on both sides to compare against. ## First derivative test URL: https://www.calclearn.org/glossary/first-derivative-test Last updated: 2026-07-28 Answer first: The first derivative test classifies a critical point by examining the sign of the derivative on both sides. A change from positive to negative gives a local maximum, negative to positive gives a local minimum, and no sign change means neither. The practical method is a sign chart: mark every critical point and every point where $f'$ is undefined on a number line, test one value in each resulting interval, and record the sign of $f'$ there. Unlike the second derivative test, this one never fails to give an answer, and it works at points where the derivative does not exist. That reliability is why it is the safer default. NOTE (Justify with the sign change): On free response, the credited justification names the sign change of $f'$, not the value of $f$. Writing that $f'$ changes from positive to negative at $x = 2$ is what earns the point. ## Fundamental Theorem of Calculus Part 1 URL: https://www.calclearn.org/glossary/ftc-part-1 Last updated: 2026-07-28 Answer first: Part 1 of the Fundamental Theorem of Calculus says that differentiating an integral with a variable upper bound returns the integrand evaluated at that bound. It is the statement that differentiation and integration are inverse operations. Also called: FTC Part 1 $$\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$$ The lower bound being constant is what makes this work; it contributes nothing to the derivative. If the upper bound is a function of $x$, the chain rule adds a factor of its derivative. NOTE (The mistake): Forgetting the chain rule factor. The derivative of $\int_0^{x^2} f(t)\,dt$ is $f(x^2) \cdot 2x$, not $f(x^2)$. ## Fundamental Theorem of Calculus Part 2 URL: https://www.calclearn.org/glossary/ftc-part-2 Last updated: 2026-07-28 Answer first: Part 2 of the Fundamental Theorem of Calculus says a definite integral equals any antiderivative evaluated at the upper bound minus the same antiderivative at the lower bound. It is what makes definite integrals computable without Riemann sums. Also called: FTC Part 2, Evaluation theorem $$\int_a^b f(x)\,dx = F(b) - F(a), \quad \text{where } F' = f$$ Any antiderivative works, because the constant of integration cancels in the subtraction. That is why no plus C appears in a definite integral. The theorem requires $f$ to be continuous on the closed interval. Applying it blindly across a vertical asymptote produces nonsense, sometimes even a negative value for an obviously positive area. NOTE (Change the bounds with the variable): After a substitution you must either convert the bounds to the new variable or convert back before evaluating. Mixing old bounds with a new variable is a routine and costly error. ## General solution URL: https://www.calclearn.org/glossary/general-solution Last updated: 2026-07-28 Answer first: The general solution of a differential equation is the entire family of functions that satisfy it, expressed with an arbitrary constant. Every specific solution is obtained by choosing a value for that constant. Because the constant is unspecified, the general solution describes infinitely many curves at once, which is precisely what a slope field draws. Any one of them is a particular solution. NOTE (Solve for y): An answer left in implicit form is usually incomplete on the AP exam. Unless the question says otherwise, isolate $y$ before finishing. ## Geometric series URL: https://www.calclearn.org/glossary/geometric-series Last updated: 2026-07-28 Answer first: A geometric series is one where each term is a fixed multiple of the previous term. It converges exactly when the absolute value of that common ratio is less than one, and then its sum is the first term divided by one minus the ratio. Also called: Common ratio $$\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}, \qquad |r| < 1$$ It is one of the very few series whose exact sum you can write down, which makes it the workhorse for comparison arguments and for building power series. NOTE (The mistake): Using the wrong first term. In the formula $a$ is the first term actually present in the series, so a sum starting at $n = 3$ has $a = ar^3$ from the original pattern, not the $n = 0$ term. ## Harmonic series URL: https://www.calclearn.org/glossary/harmonic-series Last updated: 2026-07-28 Answer first: The harmonic series is the sum of the reciprocals of the positive integers. Its terms shrink to zero, yet the series diverges, which makes it the standard proof that shrinking terms are not enough for convergence. $$\sum_{n=1}^{\infty} \frac{1}{n} = \infty$$ It is the p-series with $p = 1$, sitting exactly at the boundary between convergence and divergence, which is why it comes up constantly in comparison tests. The alternating harmonic series does converge, to $\ln 2$. It is the classic example of conditional convergence: convergent as written, divergent once you take absolute values. ## Higher-order derivative URL: https://www.calclearn.org/glossary/higher-order-derivative Last updated: 2026-07-28 Answer first: A higher-order derivative is what you get by differentiating repeatedly. The second derivative is the derivative of the derivative and measures how the rate of change is itself changing, which is what determines concavity and acceleration. Also called: Second derivative $$f''(x) = \frac{d}{dx}\left[f'(x)\right]$$ In motion problems the chain of meanings is concrete: position, then velocity, then acceleration. In graph analysis, the first derivative controls increasing and decreasing behavior while the second controls concavity. Notation varies by order. Primes are standard through the third derivative, after which a parenthesized superscript takes over, as in $f^{(4)}(x)$. In Leibniz form the second derivative is $\frac{d^2y}{dx^2}$. NOTE (The mistake): Reading $\frac{d^2y}{dx^2}$ as something squared. It is a single symbol for differentiating twice, not a power of anything. ## Horizontal asymptote URL: https://www.calclearn.org/glossary/horizontal-asymptote Last updated: 2026-07-28 Answer first: A horizontal asymptote is a horizontal line that a graph approaches as the input grows without bound in either direction. A function has one exactly when its limit at infinity or at negative infinity is a finite number. Because it is defined by a limit at infinity, a function can have at most two horizontal asymptotes, one in each direction. Rational functions have the same one on both sides, but $\arctan x$ and functions built from $\sqrt{x^2}$ often differ. NOTE (Crossing is allowed): Nothing stops a graph from crossing its horizontal asymptote, even many times. The asymptote describes only the long-run behavior, not a barrier. ## Implicit function URL: https://www.calclearn.org/glossary/implicit-function Last updated: 2026-07-28 Answer first: An implicit function is one defined by an equation relating the variables without the output isolated, such as the equation of a circle. Its derivative is found by differentiating both sides and solving for the derivative rather than by rearranging first. Many implicit equations do not define a single function at all. A circle fails the vertical line test, yet it still has a well-defined tangent slope at almost every point, which is what implicit differentiation computes. The resulting derivative usually depends on both variables, so evaluating it requires a full point rather than just an $x$ value. ## Improper integral URL: https://www.calclearn.org/glossary/improper-integral Last updated: 2026-07-28 Answer first: An improper integral is one with an infinite bound of integration or an integrand that grows without bound somewhere on the interval. It is defined as a limit of ordinary definite integrals, and it converges if that limit is finite and diverges otherwise. $$\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx$$ The procedure is always the same: replace the problematic bound with a variable, evaluate the resulting definite integral, then take the limit. Skipping the limit notation costs credit even when the arithmetic is right. If the integrand is unbounded at an interior point, split the integral there and treat each piece separately. Both pieces must converge for the whole integral to converge. NOTE (The mistake): Substituting infinity directly as if it were a number. Infinity is not a value you can plug in, and the limit is the entire content of the definition. ## Increasing function URL: https://www.calclearn.org/glossary/increasing-function Last updated: 2026-07-28 Answer first: A function is increasing on an interval where its derivative is positive and decreasing where its derivative is negative. Increasing behaviour is a property of an interval, not of a single point. Also called: Decreasing function, Monotonic The method is a sign chart for $f'$, using the critical points as dividers. Report the answer as intervals of $x$ values, since that is what the definition describes. NOTE (The mistake): Reporting where the function is increasing using $y$ values or naming a single point. The answer to an increasing or decreasing question is always a set of intervals on the $x$ axis. ## Indefinite integral URL: https://www.calclearn.org/glossary/indefinite-integral Last updated: 2026-07-28 Answer first: An indefinite integral is the collection of all antiderivatives of a function, written with an integral sign and no limits. Its result is a family of functions plus a constant of integration, which is what distinguishes it from a definite integral. $$\int f(x)\,dx = F(x) + C$$ The difference from a definite integral is a difference of type: an indefinite integral is a function, a definite integral is a number. That single distinction resolves most confusion about when the constant of integration is needed. NOTE (The mistake): Leaving off the constant of integration. Without it the answer names one antiderivative rather than all of them, and it is a routine deduction on free response. ## Indeterminate form URL: https://www.calclearn.org/glossary/indeterminate-form Last updated: 2026-07-28 Answer first: An indeterminate form is a limit expression whose value cannot be determined from its form alone, such as zero over zero or infinity minus infinity. It is not an answer but a signal that you need algebra or L'Hopital's rule to resolve the limit. Also called: 0/0 The seven indeterminate forms are $\frac{0}{0}$, $\frac{\infty}{\infty}$, $0 \cdot \infty$, $\infty - \infty$, $1^{\infty}$, $0^0$, and $\infty^0$. Each can produce any value depending on the specific functions involved, which is exactly why the form alone decides nothing. Forms like $\frac{1}{0}$ and $\frac{\infty}{0}$ are not indeterminate. They tell you the limit is infinite, which is a definite conclusion about how the limit fails to exist. NOTE (The mistake): Writing $\frac{0}{0} = 1$ or $\frac{0}{0} = 0$. Getting an indeterminate form means you have learned nothing yet, not that you have found the answer. ## Infinite discontinuity URL: https://www.calclearn.org/glossary/infinite-discontinuity Last updated: 2026-07-28 Answer first: An infinite discontinuity occurs where at least one one-sided limit is infinite, so the function grows without bound near the point. It always corresponds to a vertical asymptote and can never be removed by redefining a value. Also called: Essential discontinuity These arise when a denominator approaches zero and the numerator does not. In $\frac{1}{(x-3)^2}$ the function goes to $+\infty$ on both sides of 3; in $\frac{1}{x-3}$ it goes to $-\infty$ from the left and $+\infty$ from the right. NOTE (Check before concluding): A zero denominator does not guarantee an infinite discontinuity. If the numerator shares the factor, the discontinuity may be removable instead, so always factor first. ## Infinite limit URL: https://www.calclearn.org/glossary/infinite-limit Last updated: 2026-07-28 Answer first: An infinite limit means the function grows without bound as the input approaches a point, written as equal to positive or negative infinity. Because infinity is not a number, an infinite limit is a description of how the limit fails to exist, not a value. $$\lim_{x \to a} f(x) = \infty$$ Writing $\lim_{x \to a} f(x) = \infty$ is shorthand for a specific kind of failure: the outputs exceed every bound you name. It tells you more than simply saying the limit does not exist, which is why the notation is worth keeping. An infinite limit at $x = a$ means the line $x = a$ is a vertical asymptote. Check each side separately: a function can go to $+\infty$ on one side and $-\infty$ on the other, as $\frac{1}{x}$ does at 0. ## Inflection point URL: https://www.calclearn.org/glossary/inflection-point Last updated: 2026-07-28 Answer first: An inflection point is a point on a graph where the concavity changes from up to down or from down to up. Finding one requires the second derivative to change sign there, which is a stronger condition than simply equalling zero. Also called: Point of inflection The procedure is to find where $f''$ is zero or undefined, then build a sign chart for $f''$ and confirm an actual change of sign. The point must also be in the domain of $f$. At an inflection point the first derivative reaches a local maximum or minimum, so it is where the graph is steepest or shallowest locally. In applied contexts it is the moment a rate stops growing and starts shrinking. NOTE (The mistake): Assuming $f''(c) = 0$ makes $c$ an inflection point. For $f(x) = x^4$ the second derivative is zero at the origin but stays positive on both sides, so the concavity never changes. ## Integral test URL: https://www.calclearn.org/glossary/integral-test Last updated: 2026-07-28 Answer first: The integral test says that for a positive, decreasing, continuous function, the series of its values and the improper integral of the function either both converge or both diverge. It is what proves the p-series rule. All three conditions must be checked and stated: the function must be positive, decreasing, and continuous on the interval of interest. Skipping the verification is a common way to lose credit even with a correct conclusion. NOTE (The integral is not the sum): The test tells you whether the series converges, not what it converges to. The value of the integral is generally not the sum of the series. ## Integrand URL: https://www.calclearn.org/glossary/integrand Last updated: 2026-07-28 Answer first: The integrand is the function being integrated, written between the integral sign and the differential. Identifying it clearly is what tells you which integration technique applies. In $\int_1^4 x^2\,dx$ the integrand is $x^2$, the bounds are 1 and 4, and $dx$ names the variable of integration. Each piece answers a different question about the problem. Choosing a method is really a matter of reading the integrand's structure: a composition suggests substitution, a product of unlike types suggests integration by parts, and a rational function suggests partial fractions. ## Intermediate Value Theorem URL: https://www.calclearn.org/glossary/intermediate-value-theorem Last updated: 2026-07-28 Answer first: The Intermediate Value Theorem says that if a function is continuous on a closed interval, it attains every value between the two endpoint values somewhere on that interval. It is the standard tool for proving a solution exists without finding it. Also called: IVT $$f \text{ continuous on } [a,b], \ N \text{ between } f(a) \text{ and } f(b) \implies \exists c \in (a,b), \ f(c) = N$$ To show an equation has a root, rewrite it as $f(x) = 0$, verify $f$ is continuous on a closed interval, and show $f(a)$ and $f(b)$ have opposite signs. Zero lies between them, so the IVT guarantees a $c$ with $f(c) = 0$. For full credit you must state continuity explicitly and show the two endpoint values. Naming the theorem without checking its hypothesis is the most common way to lose the point. NOTE (What it does not say): The IVT proves existence only. It never tells you how many solutions there are or where they are, and it says nothing when the function is discontinuous. ## Interval of convergence URL: https://www.calclearn.org/glossary/interval-of-convergence Last updated: 2026-07-28 Answer first: The interval of convergence is the complete set of inputs for which a power series converges. You find the radius with the ratio test, then test each endpoint separately, because the ratio test is inconclusive exactly there. Endpoint testing is a separate job requiring a different test, usually the alternating series test on one side and a p-series comparison on the other. All four bracket combinations are possible. NOTE (The mistake): Reporting the radius and stopping, or assuming both endpoints behave the same way. They very often do not, which is why each is checked on its own. ## Inverse function derivative URL: https://www.calclearn.org/glossary/inverse-function-derivative Last updated: 2026-07-28 Answer first: The derivative of an inverse function at a point equals the reciprocal of the original function's derivative, evaluated at the matching point. You never need a formula for the inverse itself, only the point correspondence. $$\left(f^{-1}\right)'(b) = \frac{1}{f'\left(f^{-1}(b)\right)}$$ The geometry is the reason: reflecting a graph across the line $y = x$ swaps rise and run, so slopes invert. If $f$ passes through $(2, 7)$ with slope 4, then $f^{-1}$ passes through $(7, 2)$ with slope $\frac{1}{4}$. The reliable procedure is to find the input $a$ with $f(a) = b$, compute $f'(a)$, and take its reciprocal. Almost all the difficulty in these problems is bookkeeping about which point goes where. NOTE (The mistake): Evaluating $f'$ at $b$ instead of at $f^{-1}(b)$. The inner evaluation point belongs to the original function's domain, not the inverse's. ## Jump discontinuity URL: https://www.calclearn.org/glossary/jump-discontinuity Last updated: 2026-07-28 Answer first: A jump discontinuity occurs at a point where both one-sided limits exist but disagree, so the graph steps abruptly from one value to another. Because the one-sided limits differ, the two-sided limit does not exist and no redefinition can repair the break. Also called: Step discontinuity Jump discontinuities are the signature of piecewise functions whose pieces do not meet, and of step functions such as the greatest integer function. To find one, evaluate both one-sided limits and compare. A common exam task is to choose a constant that makes a piecewise function continuous. Set the left-hand limit equal to the right-hand limit and solve, then confirm the shared value also equals the function value at the seam. ## Lagrange error bound URL: https://www.calclearn.org/glossary/lagrange-error-bound Last updated: 2026-07-28 Answer first: The Lagrange error bound limits how far a Taylor polynomial can be from the function it approximates. It uses the maximum size of the next derivative on the interval between the centre and the point of interest. Also called: Taylor remainder $$\left|R_n(x)\right| \le \frac{\max\left|f^{(n+1)}(z)\right|}{(n+1)!}\left|x-a\right|^{n+1}$$ It has the shape of the next term in the series, with the derivative replaced by its largest possible value on the interval. That substitution is what turns an unknown remainder into a computable bound. For sine and cosine every derivative is bounded by one, so the maximum is simply one and the bound reduces to a factorial expression. NOTE (It is a bound, not the error): The actual error is usually much smaller. Questions ask you to show the error is less than some amount, which the bound does even when it is loose. ## Leibniz notation URL: https://www.calclearn.org/glossary/leibniz-notation Last updated: 2026-07-28 Answer first: Leibniz notation writes a derivative as dy over dx, which names both the changing quantity and the variable it changes with respect to. That explicitness makes it the standard notation for related rates, implicit differentiation, and any problem with several variables. Also called: dy/dx $$\frac{dy}{dx}$$ Prime notation is more compact and reads well for a single named function, as in $f'(x)$. Leibniz notation carries more information and is what you want the moment more than one variable is in play. The convention on this site follows the exam: primes for named functions, Leibniz for implicit differentiation and related rates, and never both inside one worked example. NOTE (Not a fraction): Treating $\frac{dy}{dx}$ as a genuine quotient is informal shorthand. It behaves like one in separable differential equations and substitution, but the symbol is defined as a limit. ## Limit URL: https://www.calclearn.org/glossary/limit Last updated: 2026-07-28 Answer first: A limit is the single value a function approaches as its input approaches a given number. The limit describes where the function is heading, not where it lands, so a limit can exist at a point where the function is undefined. Also called: Limit of a function $$\lim_{x \to a} f(x) = L$$ Reading $\lim_{x \to a} f(x) = L$ out loud: as $x$ gets arbitrarily close to $a$ from either side, $f(x)$ gets arbitrarily close to $L$. The value $f(a)$ never enters into it. That is the whole point of a limit, and it is what makes limits the right tool for defining a derivative, where the interesting quantity is undefined at the exact point you care about. For the limit to exist, both one-sided limits must exist and agree. If the function approaches 3 from the left and 5 from the right, no single value describes where it is heading, so the limit does not exist. NOTE (The mistake): Assuming $\lim_{x \to a} f(x)$ always equals $f(a)$. That is true only when $f$ is continuous at $a$, which is a stronger condition, not the definition. ## Limit at infinity URL: https://www.calclearn.org/glossary/limit-at-infinity Last updated: 2026-07-28 Answer first: A limit at infinity is the value a function approaches as its input grows without bound in the positive or negative direction. It describes the end behavior of the graph, and a finite limit at infinity is exactly what produces a horizontal asymptote. Also called: End behavior $$\lim_{x \to \infty} f(x) = L$$ For a rational function, compare degrees. If the numerator degree is smaller, the limit is 0. If the degrees are equal, the limit is the ratio of leading coefficients. If the numerator degree is larger, the function grows without bound and there is no finite limit. The reliable general method is to divide the numerator and the denominator by the highest power of $x$ in the denominator, then use the fact that $\frac{1}{x^n} \to 0$ as $x \to \infty$ for any positive $n$. NOTE (Watch the negative direction): As $x \to -\infty$, $\sqrt{x^2}$ equals $|x| = -x$, not $x$. Forgetting this flips the sign of the answer and is the single most common error on end-behavior problems involving square roots. ## Limit comparison test URL: https://www.calclearn.org/glossary/limit-comparison-test Last updated: 2026-07-28 Answer first: The limit comparison test compares two series of positive terms by taking the limit of their ratio. If that limit is finite and positive, both series converge or both diverge together. It is easier to apply than direct comparison because no inequality is needed, only that the terms grow at comparable rates. Choose the comparison series by keeping the dominant power in the numerator and denominator. For $\sum \frac{3n^2 + 1}{n^4 - 2}$, the dominant behaviour is $\frac{n^2}{n^4} = \frac{1}{n^2}$, so compare with that convergent p-series. NOTE (The limit must be finite and positive): A limit of zero or infinity makes the standard version inconclusive, so pick a comparison series that matches the growth rate more closely. ## Limit laws URL: https://www.calclearn.org/glossary/limit-laws Last updated: 2026-07-28 Answer first: The limit laws let you distribute a limit across sums, differences, products, quotients, and powers. Each law requires the individual limits to exist, and the quotient law additionally requires the denominator's limit to be nonzero. These are what justify direct substitution for continuous functions. Because polynomials are built from sums and products of continuous pieces, their limits can always be evaluated by plugging in. NOTE (The conditions are not decoration): Splitting a limit when one piece does not exist can produce a wrong answer. Applied to an indeterminate form, the quotient law simply does not apply. ## Linearization URL: https://www.calclearn.org/glossary/linearization Last updated: 2026-07-28 Answer first: Linearization approximates a function near a point by using its tangent line there. Because a smooth curve and its tangent line are nearly identical close to the point of tangency, the line gives a good estimate for inputs near that point. Also called: Tangent line approximation, Local linear approximation $$L(x) = f(a) + f'(a)(x - a)$$ The whole method is choosing a nearby point $a$ where $f(a)$ and $f'(a)$ are easy. To estimate $\sqrt{4.1}$, take $a = 4$, where the value is 2 and the derivative is $\frac{1}{4}$, giving $2 + \frac{1}{4}(0.1) = 2.025$. Concavity tells you the direction of the error. If the graph is concave down the tangent line sits above the curve, so the estimate is too large; if concave up, the estimate is too small. NOTE (The mistake): Choosing $a$ far from the input you care about. Accuracy falls off quickly with distance, and an inconvenient $a$ defeats the purpose of the method. ## Logarithmic differentiation URL: https://www.calclearn.org/glossary/logarithmic-differentiation Last updated: 2026-07-28 Answer first: Logarithmic differentiation is a technique where you take the natural logarithm of both sides of an equation before differentiating implicitly. It converts products into sums and exponents into coefficients, which is what makes a variable base raised to a variable power differentiable. It is the only reasonable route for something like $y = x^x$, where neither the power rule nor the exponential rule applies because the variable appears in both the base and the exponent. The steps are fixed: take $\ln$ of both sides, use log properties to break the expression apart, differentiate implicitly, then multiply through by $y$ and substitute the original expression back in. NOTE (Do not forget the last step): Implicit differentiation leaves you with $\frac{dy}{dx}$ in terms of $y$. The answer is not finished until you replace $y$ with the original function of $x$. ## Logistic growth URL: https://www.calclearn.org/glossary/logistic-growth Last updated: 2026-07-28 Answer first: Logistic growth models a quantity that grows nearly exponentially when small but levels off as it approaches a maximum called the carrying capacity. Its growth rate is fastest when the quantity is exactly half the carrying capacity. Also called: Carrying capacity $$\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)$$ The factor $\left(1 - \frac{P}{L}\right)$ is the brake. When $P$ is small it is close to 1 and growth looks exponential; as $P$ approaches $L$ it approaches zero and growth stalls. The solution curve is S-shaped, with an inflection point at $P = \frac{L}{2}$. That inflection point is where the population is growing fastest, which is the fact most questions turn on. NOTE (You rarely need the solved form): Most exam questions ask for the carrying capacity, the limiting behaviour, or where growth is fastest. All three come straight from the differential equation without solving it. ## Maclaurin series URL: https://www.calclearn.org/glossary/maclaurin-series Last updated: 2026-07-28 Answer first: A Maclaurin series is a Taylor series centred at zero. It is not a different object, just the most common special case, and the standard ones for the exponential, sine, and cosine functions are worth knowing by heart. $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$ The four to memorize are $e^x = \sum \frac{x^n}{n!}$, $\sin x = \sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}$, $\cos x = \sum \frac{(-1)^n x^{2n}}{(2n)!}$, and $\frac{1}{1-x} = \sum x^n$ for $|x| < 1$. Almost every series question builds from those four by substitution, multiplication by a power, differentiation, or integration, which is far faster than computing derivatives one at a time. ## Mean Value Theorem URL: https://www.calclearn.org/glossary/mean-value-theorem Last updated: 2026-07-28 Answer first: The Mean Value Theorem says that if a function is continuous on a closed interval and differentiable on the open interval, then somewhere inside there is a point where the instantaneous rate of change equals the average rate of change across the whole interval. Also called: MVT $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ Geometrically, some tangent line is parallel to the secant line joining the endpoints. In a motion context it says that if you average 60 miles per hour over a trip, at some instant your speedometer read exactly 60. The two hypotheses differ deliberately: continuity is required on the closed interval including endpoints, while differentiability is only required on the open interior. Stating both is required for credit. NOTE (The mistake): Applying it to a function with a corner inside the interval. The absolute value function on $[-1, 1]$ has average rate zero but no point where the derivative is zero, because it is not differentiable at the origin. ## Net change theorem URL: https://www.calclearn.org/glossary/net-change-theorem Last updated: 2026-07-28 Answer first: The net change theorem says that integrating a rate of change over an interval gives the net change in the underlying quantity. It is the interpretation of the Fundamental Theorem of Calculus that applied problems actually use. $$\int_a^b F'(t)\,dt = F(b) - F(a)$$ This is what turns a rate into a total. If water flows into a tank at $r(t)$ litres per minute, then $\int_0^{10} r(t)\,dt$ is the number of litres added over those ten minutes, not the flow rate at any instant. To get the final amount rather than the change, add the starting amount: $F(b) = F(a) + \int_a^b F'(t)\,dt$. That form answers most tank and population free-response questions directly. NOTE (Units): The integral of a rate has units of rate times the variable of integration, so litres per minute integrated over minutes gives litres. Checking units catches setup errors quickly. ## Normal line URL: https://www.calclearn.org/glossary/normal-line Last updated: 2026-07-28 Answer first: The normal line at a point on a curve is the line through that point perpendicular to the tangent line. Its slope is the negative reciprocal of the derivative there, provided the derivative is not zero. $$y - f(a) = -\frac{1}{f'(a)}(x - a)$$ Everything about the tangent line carries over except the slope, which flips and changes sign. If the tangent slope is $\frac{2}{3}$, the normal slope is $-\frac{3}{2}$. NOTE (Two special cases): If $f'(a) = 0$ the tangent is horizontal, so the normal line is vertical with equation $x = a$. If the tangent is vertical, the normal line is horizontal. ## nth term test URL: https://www.calclearn.org/glossary/nth-term-test Last updated: 2026-07-28 Answer first: The nth term test says that if the terms of a series do not approach zero, the series diverges. It can only ever prove divergence, because terms approaching zero says nothing about whether the series converges. Also called: Divergence test, Test for divergence $$\lim_{n \to \infty} a_n \ne 0 \implies \sum a_n \ \text{diverges}$$ It is the cheapest test, so it is worth running first. If the limit of the terms is anything other than zero, you are done in one line. NOTE (The mistake): Concluding convergence when the limit is zero. The harmonic series has terms going to zero and diverges anyway, so a zero limit means the test is inconclusive and you need another one. ## Objective function URL: https://www.calclearn.org/glossary/objective-function Last updated: 2026-07-28 Answer first: The objective function in an optimization problem is the quantity being maximized or minimized. The constraint is a separate equation relating the variables, used to rewrite the objective in terms of a single variable before differentiating. Also called: Constraint Telling them apart is the whole setup. The objective is whatever the question asks you to make largest or smallest; the constraint is the fixed condition, such as a given perimeter or a fixed volume. NOTE (The mistake): Differentiating the constraint instead of the objective. The constraint exists only to eliminate a variable; the derivative you set to zero belongs to the objective. ## One-sided limit URL: https://www.calclearn.org/glossary/one-sided-limit Last updated: 2026-07-28 Answer first: A one-sided limit is the value a function approaches as the input comes in from one direction only. The left-hand limit uses inputs below the point and the right-hand limit uses inputs above it. The two-sided limit exists exactly when both agree. Also called: Left-hand limit, Right-hand limit $$\lim_{x \to a^-} f(x) \quad \text{and} \quad \lim_{x \to a^+} f(x)$$ The notation $x \to a^-$ means $x$ approaches $a$ through values less than $a$, and $x \to a^+$ means through values greater than $a$. The superscript describes which side of the number line you are on, not the sign of the function. One-sided limits are the tool for piecewise functions, jump discontinuities, and any question about behavior at an endpoint of a domain. They are also how you check differentiability at a corner. NOTE (The mistake): Reading the minus in $a^-$ as a negative number. In $\lim_{x \to 3^-}$ the inputs are values like 2.9 and 2.99, all positive, all just below 3. ## p-series URL: https://www.calclearn.org/glossary/p-series Last updated: 2026-07-28 Answer first: A p-series is the sum of the reciprocals of the positive integers raised to a fixed power p. It converges when p is greater than one and diverges when p is less than or equal to one. $$\sum_{n=1}^{\infty} \frac{1}{n^p} \ \text{converges} \iff p > 1$$ The rule follows from the integral test applied to $\frac{1}{x^p}$, which is why the boundary sits exactly at $p = 1$: that is where the antiderivative switches from a power to a logarithm. NOTE (The mistake): Including $p = 1$ as convergent. The harmonic series is the $p = 1$ case and diverges, so the inequality is strict. ## Parametric equations URL: https://www.calclearn.org/glossary/parametric-equations Last updated: 2026-07-28 Answer first: Parametric equations define the coordinates of a point separately as functions of a third variable called the parameter, usually time. They describe not just the shape of a curve but the direction and speed with which it is traced. Also called: Parameter $$x = x(t), \quad y = y(t)$$ The parameter buys you curves that are not functions of $x$, such as circles, and it records the traversal. A curve traced clockwise and the same curve traced counterclockwise are different parametrizations. The slope comes from the chain rule as $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, valid wherever $\frac{dx}{dt}$ is not zero. The second derivative differentiates that quotient with respect to $t$, then divides by $\frac{dx}{dt}$ again. NOTE (The mistake): Computing the second derivative as $\frac{d^2y/dt^2}{d^2x/dt^2}$. That is not the chain rule and gives the wrong answer. ## Partial fractions URL: https://www.calclearn.org/glossary/partial-fractions Last updated: 2026-07-28 Answer first: Partial fraction decomposition rewrites a rational function as a sum of simpler fractions with linear denominators. Each piece then integrates to a natural logarithm, which is what makes otherwise impossible rational integrals routine. Also called: Partial fraction decomposition On the AP exam the denominator always factors into distinct linear factors, so the setup is $\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$. Clear the denominators and solve for the constants. The fastest way to find the constants is to substitute the values that make each factor vanish, which knocks out all but one unknown at a time. NOTE (Check the degree first): Partial fractions require the numerator degree to be lower than the denominator degree. If it is not, do polynomial long division before decomposing. ## Partial sum URL: https://www.calclearn.org/glossary/partial-sum Last updated: 2026-07-28 Answer first: The nth partial sum of a series is the total of its first n terms. The series converges precisely when this sequence of partial sums approaches a finite limit, which is what makes an infinite sum meaningful. $$S_N = \sum_{n=1}^{N} a_n$$ Partial sums turn an infinite question into a sequence question. For a geometric series they have a closed form, which is how the geometric sum formula is derived. For an alternating series the partial sums bracket the true sum, jumping above and below it. That is exactly why the alternating series error bound is so simple. ## Particular solution URL: https://www.calclearn.org/glossary/particular-solution Last updated: 2026-07-28 Answer first: A particular solution is the one member of a differential equation's solution family that passes through a specified point. The initial condition is what determines the arbitrary constant and selects that single curve. Also called: Initial value problem The order matters. Find the general solution first, substitute the initial condition to solve for the constant, then write the final answer with that value in place. NOTE (Substitute early when separating): In a separable equation it is often cleaner to apply the initial condition right after integrating, before rearranging, because the constant is easier to isolate at that stage. ## Polar coordinates URL: https://www.calclearn.org/glossary/polar-coordinates Last updated: 2026-07-28 Answer first: Polar coordinates locate a point by its distance from the origin and the angle it makes with the positive horizontal axis. A polar curve expresses that distance as a function of the angle, which makes circles and roses far simpler to describe than in rectangular form. Also called: Polar curve $$x = r\cos\theta, \quad y = r\sin\theta$$ Polar curves are a special case of parametric equations with $\theta$ as the parameter, which is why the slope formula is the parametric one applied to $x = r(\theta)\cos\theta$ and $y = r(\theta)\sin\theta$. Area in polar coordinates is not the usual integral of a height. Sectors, not rectangles, are the natural slice, giving $A = \frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$. NOTE (The mistake): Forgetting the one half or the square in the polar area formula. Both come from the area of a circular sector and neither is optional. ## Power rule URL: https://www.calclearn.org/glossary/power-rule Last updated: 2026-07-28 Answer first: The power rule says that to differentiate a power of the variable, multiply by the exponent and then reduce the exponent by one. It holds for every real exponent, including negative and fractional ones. $$\frac{d}{dx}\left[x^n\right] = n x^{n-1}$$ Its reach is wider than it first appears, because roots and reciprocals are powers in disguise. Writing $\sqrt{x}$ as $x^{1/2}$ and $\frac{1}{x^3}$ as $x^{-3}$ turns both into one-step problems. NOTE (The mistake): Applying it to $a^x$, where the variable is the exponent rather than the base. That is an exponential function and needs a different rule entirely. ## Power series URL: https://www.calclearn.org/glossary/power-series Last updated: 2026-07-28 Answer first: A power series is an infinite sum of constant multiples of powers of the variable minus a fixed centre. It converges for the centre always, and generally on an interval around it whose half-width is the radius of convergence. $$\sum_{n=0}^{\infty} c_n (x - a)^n$$ The value of $x$ decides everything, so a power series is really a family of series. Finding where it converges is the first question asked about one. Inside the interval of convergence a power series can be differentiated and integrated term by term, and the radius stays the same, though the endpoint behaviour can change. ## Product rule URL: https://www.calclearn.org/glossary/product-rule Last updated: 2026-07-28 Answer first: The product rule differentiates a product of two functions: multiply the first by the derivative of the second, then add the second times the derivative of the first. The derivative of a product is never the product of the derivatives. $$\frac{d}{dx}\left[f(x)g(x)\right] = f(x)g'(x) + g(x)f'(x)$$ The rule is needed only when both factors contain the variable. If one factor is a constant, the constant multiple rule applies and the work is much shorter. NOTE (The mistake): Writing the derivative of a product as $f'(x)g'(x)$. Testing it on $x \cdot x = x^2$ shows the error immediately: the correct answer is $2x$, not 1. ## Properties of definite integrals URL: https://www.calclearn.org/glossary/definite-integral-properties Last updated: 2026-07-28 Answer first: Definite integrals obey a small set of rules: swapping the bounds reverses the sign, equal bounds give zero, constants factor out, sums split apart, and an integral can be broken at any interior point. $$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$$ The splitting rule holds for any $c$, including values outside the interval, which is what lets you combine and rearrange given integrals on table-based free-response questions. Sign reversal on swapping bounds follows from the Fundamental Theorem: $F(b) - F(a)$ is the negative of $F(a) - F(b)$. NOTE (No product rule): There is no rule turning the integral of a product into a product of integrals. That mistake has no analogue that works. ## Quotient rule URL: https://www.calclearn.org/glossary/quotient-rule Last updated: 2026-07-28 Answer first: The quotient rule differentiates a fraction: the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the denominator squared. The order matters because of the subtraction. $$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{\left(g(x)\right)^2}$$ Unlike the product rule, the quotient rule is not symmetric, so swapping the two terms in the numerator flips the sign of the whole answer. That is the single most common error with it. NOTE (Often avoidable): When the denominator is a single power, rewriting as a negative exponent and using the power rule is faster and less error-prone than the quotient rule. ## Radius of convergence URL: https://www.calclearn.org/glossary/radius-of-convergence Last updated: 2026-07-28 Answer first: The radius of convergence is the distance from the centre of a power series to the edge of the region where it converges. It is found by applying the ratio test and solving the resulting inequality for the variable. Applying the ratio test gives a limit involving the absolute value of $x - a$. Setting that limit less than one produces an inequality of the form $|x - a| < R$, and $R$ is the radius. The two extremes are real: a radius of zero means the series converges only at its centre, and an infinite radius means it converges for every real number, as the series for $e^x$ does. NOTE (Radius is not the interval): The radius is a single number. The interval of convergence additionally requires testing both endpoints separately, since the ratio test is silent there. ## Related rate URL: https://www.calclearn.org/glossary/related-rate Last updated: 2026-07-28 Answer first: A related rate is the rate of change of one quantity expressed in terms of the rate of change of another that it depends on. These problems link the quantities with an equation, then differentiate both sides with respect to time. The order that keeps these problems honest is: write the geometric relationship first, differentiate with respect to time second, and substitute the given numerical values last. NOTE (The mistake): Substituting the numbers before differentiating. Doing so turns a changing quantity into a constant, whose derivative is zero, and the problem collapses. ## Removable discontinuity URL: https://www.calclearn.org/glossary/removable-discontinuity Last updated: 2026-07-28 Answer first: A removable discontinuity is a point where the limit exists but the function either is not defined there or is defined to be a different value. It appears as a hole in the graph, and redefining the single point would make the function continuous. Also called: Hole, Point discontinuity These come from a factor that cancels. In $f(x) = \frac{x^2 - 4}{x - 2}$, the numerator factors as $(x-2)(x+2)$, so the function equals $x + 2$ everywhere except $x = 2$, where it is undefined. The limit there is 4, so the hole sits at the point $(2, 4)$. The word removable is literal: define $f(2) = 4$ and the discontinuity is gone. That is impossible for a jump or an infinite discontinuity, which is what distinguishes them. NOTE (The mistake): Cancelling the factor and forgetting to exclude the point. The simplified expression $x + 2$ is a different function from the original unless you state that $x \ne 2$. ## Riemann sum URL: https://www.calclearn.org/glossary/riemann-sum Last updated: 2026-07-28 Answer first: A Riemann sum approximates a definite integral by slicing the interval into subintervals and adding up the area of a rectangle on each one. As the number of subintervals grows without bound, the sum approaches the exact value of the integral. Also called: Left Riemann sum, Right Riemann sum, Midpoint sum $$\sum_{i=1}^{n} f(x_i^*)\,\Delta x, \qquad \Delta x = \frac{b-a}{n}$$ The only choice that distinguishes the standard versions is where each rectangle takes its height: the left endpoint, the right endpoint, or the midpoint of its subinterval. The trapezoidal rule uses trapezoids instead and equals the average of the left and right sums. For an increasing function the left sum underestimates and the right sum overestimates, and for a decreasing function the roles swap. Concavity decides the trapezoidal and midpoint errors: trapezoids overestimate on a concave up curve. NOTE (The mistake): Using $n$ data points as $n$ subintervals. A table with values at 0, 2, 5, and 9 gives three subintervals of unequal width, so $\Delta x$ is not constant. ## Rolle's Theorem URL: https://www.calclearn.org/glossary/rolles-theorem Last updated: 2026-07-28 Answer first: Rolle's Theorem says that if a function is continuous on a closed interval, differentiable inside it, and takes the same value at both endpoints, then there is an interior point where the derivative is zero. It is the Mean Value Theorem with equal endpoint values. When $f(a) = f(b)$ the average rate of change is zero, so the Mean Value Theorem promises a point where $f'(c) = 0$. That is exactly a horizontal tangent somewhere between the two endpoints. The standard use is proving that a function has a critical point between two roots, which in turn bounds how many roots the function can have. ## Root test URL: https://www.calclearn.org/glossary/root-test Last updated: 2026-07-28 Answer first: The root test takes the limit of the nth root of the absolute value of the terms. A limit less than one means the series converges absolutely, greater than one means it diverges, and exactly one means the test tells you nothing. $$L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$$ It is the natural choice when the whole term is raised to the nth power, because the root cancels the exponent cleanly. For anything with factorials, the ratio test is easier. NOTE (Inconclusive at one): When the limit equals one the root test says nothing, and so does the ratio test, so you must switch to a comparison or the integral test. ## Secant line URL: https://www.calclearn.org/glossary/secant-line Last updated: 2026-07-28 Answer first: A secant line is a straight line through two points on a curve. Its slope equals the average rate of change between those points, and as the two points slide together the secant line approaches the tangent line. The secant line is the concrete object behind the definition of the derivative. Watching its slope stabilize as the second point approaches the first is the clearest way to see why the derivative is defined with a limit. You can move a point and watch this happen directly in the secant to tangent visualizer, where shrinking the gap turns the secant into the tangent in real time. ## Second derivative test URL: https://www.calclearn.org/glossary/second-derivative-test Last updated: 2026-07-28 Answer first: The second derivative test classifies a critical point by evaluating the second derivative there. A negative value means a local maximum, a positive value means a local minimum, and a value of zero makes the test inconclusive. The logic is about concavity. At a critical point on a concave down curve the graph must turn over, giving a maximum; on a concave up curve it must turn up, giving a minimum. When $f''$ is zero at the critical point the test says nothing at all, and you must fall back on the first derivative test. Compare $x^4$, $-x^4$, and $x^3$, which all have $f'(0) = f''(0) = 0$ but a minimum, a maximum, and neither. NOTE (The mistake): Concluding a point is neither a maximum nor a minimum when $f''$ is zero. Inconclusive means you need a different test, not that no extremum exists. ## Sequence URL: https://www.calclearn.org/glossary/sequence Last updated: 2026-07-28 Answer first: A sequence is an ordered list of numbers indexed by the positive integers. A series is the sum of the terms of a sequence, so a sequence is a list while a series is a total. A sequence converges when its terms approach a single limit. A series converges when its partial sums approach a limit, which is a stronger and quite different requirement. The harmonic sequence $\frac{1}{n}$ converges to 0, but the harmonic series $\sum \frac{1}{n}$ diverges. That one example is the clearest reason the distinction matters. NOTE (The mistake): Concluding a series converges because its terms go to zero. Terms going to zero is necessary but nowhere near sufficient. ## Series URL: https://www.calclearn.org/glossary/series Last updated: 2026-07-28 Answer first: A series is the sum of the terms of a sequence, often infinitely many. It converges when the sequence of its partial sums approaches a finite limit, and that limit is defined to be the sum of the series. Also called: Infinite series $$\sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} \sum_{n=1}^{N} a_n$$ Adding infinitely many numbers is not defined directly. The definition routes through partial sums, which are finite additions, and then takes a limit, so every question about a series is really a question about a limit. Deciding convergence is what the battery of tests is for, and picking the right one is most of the skill. Geometric and p-series can be settled by inspection. ## Shell method URL: https://www.calclearn.org/glossary/shell-method Last updated: 2026-07-28 Answer first: The shell method computes a volume of revolution by slicing the region parallel to the axis of rotation and treating each slice as a thin cylindrical shell. Its integrand is two pi times the radius times the height. Also called: Cylindrical shells $$V = 2\pi \int_a^b r(x)\,h(x)\,dx$$ The factor of $2\pi r$ is the circumference of the shell and $h$ is its height, so the product is the area of the unrolled rectangle. Multiplying by the thickness and integrating gives the volume. Shells are the better choice when slicing perpendicular to the axis would force you to solve the equation for the other variable, which is often messy or impossible. ## Slope field URL: https://www.calclearn.org/glossary/slope-field Last updated: 2026-07-28 Answer first: A slope field is a grid of short line segments, each drawn with the slope that a differential equation assigns to that point. It shows the shape of every solution curve at once without solving the equation. Also called: Direction field To sketch a particular solution, put your pencil on the given initial condition and follow the segments, keeping the curve parallel to the nearby marks. The curve never crosses segments, it flows along them. Matching a field to an equation is faster with structure than with arithmetic. If $\frac{dy}{dx}$ depends only on $x$, every column of segments looks identical; if it depends only on $y$, every row does. ## Solid of revolution URL: https://www.calclearn.org/glossary/solid-of-revolution Last updated: 2026-07-28 Answer first: A solid of revolution is the three-dimensional shape swept out when a plane region is rotated about a line. Its volume is computed by integrating cross-sectional areas, using the disk method, the washer method, or the shell method. The choice of method is really a choice of slice direction. Disks and washers slice perpendicular to the axis of rotation, so they integrate with respect to the axis variable; shells slice parallel to it. Use a washer rather than a disk whenever the region does not touch the axis of rotation, because the gap becomes a hole through the solid. NOTE (Radii are distances): Rotating about a line other than an axis changes every radius. The radius is always the distance from the slice to the axis, so it becomes an expression like $4 - f(x)$ rather than $f(x)$. ## Speed URL: https://www.calclearn.org/glossary/speed Last updated: 2026-07-28 Answer first: Speed is the absolute value of velocity, so it is never negative and carries no direction information. Speed increases exactly when velocity and acceleration have the same sign. $$\text{speed} = |v(t)|$$ Because speed strips the sign, a particle moving in the negative direction can still be speeding up. Questions that ask whether speed is increasing are really asking you to compare two signs. For a particle moving along a curve described parametrically, speed is the magnitude of the velocity vector, $\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$. ## Squeeze theorem URL: https://www.calclearn.org/glossary/squeeze-theorem Last updated: 2026-07-28 Answer first: The squeeze theorem says that if a function is trapped between two others near a point, and those two share the same limit there, the trapped function must have that limit too. It is the standard way to evaluate limits that resist algebra, such as x squared times sine of one over x. Also called: Sandwich theorem, Pinching theorem $$g(x) \le f(x) \le h(x) \ \text{and} \ \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \implies \lim_{x \to a} f(x) = L$$ The inequality only needs to hold near $a$, not everywhere, and it does not need to hold at $a$ itself. The theorem is useful precisely when $f$ oscillates too wildly for direct methods but is bounded by something tame. The classic application uses $-1 \le \sin\left(\frac{1}{x}\right) \le 1$, so $-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$. Both bounds go to 0 as $x \to 0$, so the middle function does too. NOTE (The mistake): Choosing bounds whose limits are different. If $g$ and $h$ do not share the same limit, the theorem says nothing at all. ## Tangent line URL: https://www.calclearn.org/glossary/tangent-line Last updated: 2026-07-28 Answer first: The tangent line to a curve at a point is the straight line through that point whose slope equals the derivative there. It is the best linear approximation of the curve near that point. $$y - f(a) = f'(a)(x - a)$$ Building the equation needs exactly two ingredients: the point $(a, f(a))$ and the slope $f'(a)$. Point-slope form assembles them directly, which is why it is the form worth memorizing. Near the point of tangency the tangent line and the curve are nearly indistinguishable. That fact is the whole content of linearization and tangent line approximation. NOTE (The mistake): Using $f'(x)$ as the slope instead of $f'(a)$. The slope of a specific tangent line is a number, obtained by evaluating the derivative at the point of tangency. ## Taylor polynomial URL: https://www.calclearn.org/glossary/taylor-polynomial Last updated: 2026-07-28 Answer first: A Taylor polynomial is the finite piece of a Taylor series obtained by stopping after a chosen degree. It approximates the function near the centre, and higher degree generally means a better approximation over a wider range. The degree counts the highest power actually present, which can be lower than the number of terms you generated. For $\sin x$ the odd powers alone appear, so the second and third degree polynomials are the same expression. The error is bounded by the Lagrange error bound, or by the first omitted term when the series alternates. NOTE (Watch which derivative): The coefficient of $(x-a)^n$ uses the nth derivative at the centre divided by $n!$. Dropping the factorial is the most common slip in building these. ## Taylor series URL: https://www.calclearn.org/glossary/taylor-series Last updated: 2026-07-28 Answer first: A Taylor series represents a function as an infinite polynomial whose coefficients come from the function's derivatives at a single centre point. Each coefficient is a derivative at the centre divided by the factorial of its order. $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$ The construction forces the polynomial to match the function's value and every derivative at the centre, which is why it approximates so well nearby and why the factorials appear. In practice you rarely differentiate repeatedly. Most series are built by substituting into, differentiating, or integrating a known series such as the ones for $e^x$, $\sin x$, $\cos x$, and $\frac{1}{1-x}$. NOTE (Centred where): A Taylor series is always about a specific centre. Changing the centre changes every coefficient, so the centre must be stated. ## Trapezoidal rule URL: https://www.calclearn.org/glossary/trapezoidal-rule Last updated: 2026-07-28 Answer first: The trapezoidal rule approximates a definite integral by joining consecutive points with straight segments and summing the resulting trapezoids. It equals the average of the left and right Riemann sums and is usually more accurate than either. Also called: Trapezoid rule $$\int_a^b f(x)\,dx \approx \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + \cdots + 2f(x_{n-1}) + f(x_n)\right]$$ The doubled interior coefficients are not arbitrary: every interior point is shared by two adjacent trapezoids, so it gets counted twice while the two endpoints are counted once. Concavity fixes the direction of the error. On a concave up curve the chords lie above the graph, so the trapezoidal rule overestimates; on a concave down curve it underestimates. NOTE (Unequal widths): The compact formula assumes equal spacing. With data at uneven intervals, compute each trapezoid separately as the average height times its own width. ## Vector-valued function URL: https://www.calclearn.org/glossary/vector-valued-function Last updated: 2026-07-28 Answer first: A vector-valued function assigns a vector to each value of a parameter, most often describing the position of a particle over time. Differentiating each component separately gives the velocity vector, and differentiating again gives acceleration. $$\mathbf{r}(t) = \langle x(t),\, y(t) \rangle$$ Everything is componentwise, which makes these easier than they look. The velocity is $\langle x'(t), y'(t) \rangle$ and the speed is its magnitude, $\sqrt{\left(x'(t)\right)^2 + \left(y'(t)\right)^2}$. Total distance travelled is the integral of speed over the time interval, while displacement is the vector difference between the ending and starting positions. NOTE (Speed is a scalar): Velocity is a vector with direction; speed is a single non-negative number. Questions asking when a particle is at rest want both velocity components to be zero at once. ## Velocity URL: https://www.calclearn.org/glossary/velocity Last updated: 2026-07-28 Answer first: Velocity is the derivative of position with respect to time. It is a signed quantity, so its sign tells you the direction of motion, and its absolute value is the speed. $$v(t) = s'(t)$$ Positive velocity means motion in the positive direction and negative velocity means motion in the negative direction. A particle is momentarily at rest when velocity equals zero, which is also where it may change direction. Integrating velocity over an interval gives displacement, the net change in position. Integrating the absolute value of velocity gives total distance travelled, which is a different and usually larger number. NOTE (The mistake): Confusing displacement with distance. If a particle moves forward 5 and back 3, the displacement is 2 but the distance travelled is 8. ## Vertical asymptote URL: https://www.calclearn.org/glossary/vertical-asymptote Last updated: 2026-07-28 Answer first: A vertical asymptote is a vertical line that a graph approaches without ever reaching, occurring wherever at least one one-sided limit is infinite. For a rational function, these sit at the zeros of the denominator that remain after all common factors cancel. The order matters: factor the numerator and denominator, cancel common factors first, and only then set the remaining denominator to zero. A factor that cancels produces a hole rather than an asymptote. A graph may cross a horizontal asymptote but can never cross a vertical one, because a vertical asymptote sits at an input where the function has no value at all. ## Vertical tangent URL: https://www.calclearn.org/glossary/vertical-tangent Last updated: 2026-07-28 Answer first: A vertical tangent occurs at a point where the derivative grows without bound with the same sign from both sides, so the tangent line is vertical. The function remains continuous there but is not differentiable, since infinite slope is not a number. The cube root function $f(x) = x^{1/3}$ has a vertical tangent at the origin. Its derivative $\frac{1}{3}x^{-2/3}$ tends to $+\infty$ from both sides, so the curve passes through smoothly while standing straight up. Vertical tangents matter for implicit differentiation, where $\frac{dy}{dx}$ often has a denominator that vanishes. Setting that denominator to zero locates the points with vertical tangents. ## Washer method URL: https://www.calclearn.org/glossary/washer-method Last updated: 2026-07-28 Answer first: The washer method computes the volume of a solid of revolution with a hole through it by subtracting the inner circle's area from the outer circle's area on every slice. It is used whenever the rotated region does not touch the axis of rotation. $$V = \pi \int_a^b \left[\left(R(x)\right)^2 - \left(r(x)\right)^2\right] dx$$ The outer radius reaches the curve farther from the axis and the inner radius reaches the nearer one. Both are distances from the axis, so rotating about a line other than an axis changes both expressions. NOTE (The mistake): Writing $(R - r)^2$ instead of $R^2 - r^2$. Those are different quantities, and only the second describes the area of an annulus. --- # Unit overviews ## Unit 1: Limits and Continuity URL: https://www.calclearn.org/units/unit-1-limits-and-continuity Last updated: 2026-07-22 Answer first: Unit 1 builds the limit: the tool for describing how a function behaves near a point it may never reach, then uses it to define continuity. It is 10-15% of the AB multiple-choice section and 5-10% of BC across 16 topics. The tested skill is picking the limit method that fits the form you get after substituting. Exam weighting: AB 10-15%, BC 5-10%. Topics: - 1.1 Introducing Calculus: Can Change Occur at an Instant? - 1.2 Defining Limits and Using Limit Notation - 1.3 Estimating Limit Values from Graphs - 1.4 Estimating Limit Values from Tables - 1.5 Determining Limits Using Algebraic Properties of Limits - 1.6 Determining Limits Using Algebraic Manipulation - 1.7 Selecting Procedures for Determining Limits - 1.8 Determining Limits Using the Squeeze Theorem - 1.9 Connecting Multiple Representations of Limits - 1.10 Exploring Types of Discontinuities - 1.11 Defining Continuity at a Point - 1.12 Confirming Continuity over an Interval - 1.13 Removing Discontinuities - 1.14 Connecting Infinite Limits and Vertical Asymptotes - 1.15 Connecting Limits at Infinity and Horizontal Asymptotes - 1.16 Working with the Intermediate Value Theorem (IVT) A limit describes what value a function approaches as the input nears a point, whether or not the function is even defined there (Topic 1.2). That gap between where a function is headed and where it actually lands is what makes calculus possible: it lets you define an instantaneous rate of change as the limit of average rates, which otherwise divide by zero at a single instant (Topic 1.1). You will express limits four ways: graphically, numerically, analytically, and verbally, but the formal epsilon-delta definition is not assessed on the AB or BC exam. | Topics | What they cover | Big idea | | --- | --- | --- | | 1.1 | Why an average rate breaks at an instant, and how a limit fixes it | Change (CHA-1) | | 1.2-1.9 | Defining, estimating, and evaluating limits, and selecting the procedure | Limits (LIM-1) | | 1.10-1.15 | Continuity, types of discontinuity, and infinite or end behavior | Limits (LIM-2) | | 1.16 | The Intermediate Value Theorem, the first existence theorem | Analysis of Functions (FUN-1) | Topic 1.7 is the heart of the unit and the reason a solver cannot do it for you: given a limit, you have to recognize which technique applies. The move is always the same. Substitute first. If you get a real number, that is the answer. If you get $\frac{0}{0}$, a common factor is hiding, so factor or multiply by a conjugate (Topic 1.6). A nonzero number over $0$ signals a vertical asymptote (Topic 1.14). A $\frac{\sin x}{x}$ shape points to a known trig limit or the squeeze theorem (Topic 1.8). Reading that returned form, not memorizing five procedures, is what the exam rewards. $$\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0$$ Continuity is where limits pay off. A function is continuous at $x = c$ when three things all hold: $f(c)$ exists, $\lim_{x \to c} f(x)$ exists, and the two are equal (Topic 1.11). When they fail, the break is one of three named types: removable, jump, or infinite (a vertical asymptote), and identifying which is a common exam task (Topic 1.10). Polynomial, rational, power, exponential, logarithmic, and trigonometric functions are continuous everywhere in their domains (Topic 1.12), which is exactly why direct substitution works for them. Unit 1 is 10-15% of the AB multiple-choice section and 5-10% of BC, and it appears on both the calculator and no-calculator parts. Expect to read limits off graphs and tables (Topics 1.3, 1.4), justify continuity at the boundary of a piecewise function (Topic 1.13), and apply the Intermediate Value Theorem. That last task tests a habit you will use all year: state and check a theorem's hypotheses before using its conclusion. The IVT guarantees a solution only after you confirm the function is continuous on a closed interval (Topic 1.16). Everything downstream rests on this unit, because every derivative and integral is itself a limit. ### How to study this unit - Run one loop on every limit: substitute first (Topic 1.5), then let the result pick the method. A real number is the answer, $\frac{0}{0}$ means factor or use a conjugate (Topic 1.6), and a nonzero number over $0$ is a vertical asymptote (Topic 1.14). That recognition is the Topic 1.7 skill the exam actually grades. - Memorize the two squeeze-theorem limits cold: $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ (Topic 1.8). They show up disguised, like $\frac{\sin 3x}{4x}$, so drill rewriting a problem into the standard shape. - Treat the continuity definition as a three-box checklist (Topic 1.11): $f(c)$ exists, the limit exists, and they match. Piecewise boundary problems (Topic 1.13) are graded on showing all three, not on the arithmetic alone. - For any Intermediate Value Theorem question (Topic 1.16), write the hypotheses before the conclusion: state that the function is continuous on the closed interval and that your target value lies between the endpoint values. No credit follows until continuity is verified. - Handle limits at infinity by comparing leading degrees (Topic 1.15) instead of plugging in a large number: the ratio of the leading terms gives the horizontal asymptote and the end behavior directly. ## Unit 2: Differentiation: Definition and Fundamental Properties URL: https://www.calclearn.org/units/unit-2-differentiation-definition Last updated: 2026-07-22 Answer first: Unit 2 turns Unit 1's limits into the derivative. It defines f'(x) as the limit of the difference quotient, ties it to the tangent-line slope, then gives the shortcut rules: power, sum, constant multiple, product, and quotient, plus derivatives of sin x, cos x, e^x, ln x, and the trig functions. Worth 10-15% of AB. Exam weighting: AB 10-15%, BC 5-10%. Topics: - 2.1 Defining Average and Instantaneous Rates of Change at a Point - 2.2 Defining the Derivative of a Function and Using Derivative Notation - 2.3 Estimating Derivatives of a Function at a Point - 2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist - 2.5 Applying the Power Rule - 2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple - 2.7 Derivatives of cos x, sin x, e^x, and ln x - 2.8 The Product Rule - 2.9 The Quotient Rule - 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions NOTE (The big idea): The derivative is a single limit wearing three costumes: the instantaneous rate of change of $f$, the slope of the line tangent to the graph, and a new function $f'(x)$ you can evaluate anywhere. Unit 2 defines that limit, then builds fast rules so you never have to compute it from scratch again. Everything starts from the difference quotient. The average rate of change of $f$ over an interval is $\frac{f(a+h)-f(a)}{h}$, equivalently $\frac{f(x)-f(a)}{x-a}$ (Topic 2.1). Let the interval shrink toward zero and, when the limit exists, you land on the instantaneous rate of change: the derivative at $a$. The same limit taken at a general input defines the derivative as a function, $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ (Topic 2.2). $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$ The unit moves in three stages. Topics 2.1 to 2.3 are about meaning: define the derivative, read it off graphs and tables, and estimate it numerically. Topic 2.4 draws the line between differentiability and continuity: differentiable forces continuous, but continuous does not force differentiable. Topics 2.5 to 2.10 are the toolkit: the power rule, the linearity rules, the named derivatives of $\sin x$, $\cos x$, $e^x$, and $\ln x$, and finally the product, quotient, and remaining trig rules. This is where method selection starts to earn points. Before you differentiate, name the structure of the expression: is it a sum of terms (differentiate term by term, Topic 2.6), a product (Topic 2.8), or a quotient (Topic 2.9)? Rewrite when it is cheaper: a radical becomes $x^{1/2}$ so the power rule applies, and the Topic 2.10 functions ($\tan x$, $\sec x$, and the rest) are easiest through their known results or by rewriting via $\sin x$ and $\cos x$. Picking the rule is the skill the exam rewards; running the rule is the easy part. | Topic | When you see | Rule | Derivative | | --- | --- | --- | --- | | 2.5 | $x^r$ | Power rule | $rx^{r-1}$ | | 2.6 | sum, difference, or $c\cdot f$ | Linearity | differentiate term by term | | 2.7 | $\sin x,\ \cos x,\ e^x,\ \ln x$ | Named results | $\cos x,\ -\sin x,\ e^x,\ \tfrac{1}{x}$ | | 2.8 | a product $fg$ | Product rule | $f'g + fg'$ | | 2.9 | a quotient $\tfrac{f}{g}$ | Quotient rule | $\tfrac{f'g - fg'}{g^2}$ | | 2.10 | $\tan x,\ \cot x,\ \sec x,\ \csc x$ | Rewrite via identities | $\sec^2 x,\ -\csc^2 x,\ \sec x\tan x,\ -\csc x\cot x$ | On the AB exam Unit 2 is 10-15% of the multiple-choice section (5-10% on BC). Expect multiple-choice items that hand you a table or graph and ask for $f'$ at a point (Topic 2.3), questions that test whether a function is differentiable at a corner or a vertical tangent (Topic 2.4), and steady computation with the product and quotient rules. Watch for a disguised derivative: a limit such as $\lim_{x \to 0} \frac{e^x - 1}{x}$ is just $f'(0)$ for $f(x)=e^x$ (Topic 2.7), so recognizing the difference-quotient pattern turns a hard limit into a one-line answer. ### How to study this unit - Learn both forms of the definition (Topics 2.1, 2.2) and recognize the pattern backward: a limit like $\lim_{x \to 0} \frac{e^x-1}{x}$ is really $f'(0)$ for $f(x)=e^x$ (Topic 2.7). - Differentiability implies continuity, but not the reverse (Topic 2.4). Memorize the two ways it fails: a corner where the left and right difference quotients disagree, like $|x|$ at $x=0$, and a vertical tangent with no slope, like $\sqrt[3]{x}$ at $x=0$. If $a$ is not in the domain of $f$, it is not in the domain of $f'$. - Classify structure before you compute. Ask whether the expression is a sum (differentiate term by term, Topic 2.6), a product (Topic 2.8), or a quotient (Topic 2.9), then apply the matching rule instead of reaching for the hardest one by reflex. - Rewrite to dodge work. For the power rule (Topic 2.5), turn radicals and reciprocals into exponents first: $\sqrt{x}=x^{1/2}$ and $\frac{1}{x^2}=x^{-2}$. For Topic 2.10, rewrite $\tan x$, $\sec x$, and the rest through $\sin x$ and $\cos x$, or use the known derivatives directly. - Practice reading derivatives off tables and graphs (Topic 2.3). From a table, estimate $f'(a)$ with a symmetric difference quotient $\frac{f(a+h)-f(a-h)}{2h}$; from a graph, read the slope of the tangent line. Remember the calculator can return a numerical derivative at a point. ## Unit 3: Differentiation: Composite, Implicit, and Inverse Functions URL: https://www.calclearn.org/units/unit-3-composite-implicit-inverse Last updated: 2026-07-22 Answer first: AP Calculus Unit 3 teaches you to differentiate functions built from other functions: composites (the chain rule), implicitly defined curves, and inverse and inverse trig functions. It is worth 5-10% of both the AB and BC exams, but the chain rule underlies nearly every derivative you take after this unit. Exam weighting: AB 5-10%, BC 5-10%. Topics: - 3.1 The Chain Rule - 3.2 Implicit Differentiation - 3.3 Differentiating Inverse Functions - 3.4 Differentiating Inverse Trigonometric Functions - 3.5 Selecting Procedures for Calculating Derivatives - 3.6 Calculating Higher-Order Derivatives Every topic in Unit 3 is the chain rule wearing a different costume. Unit 2 gave you rules for plain functions; Unit 3 handles what happens when functions are nested inside each other, tangled together implicitly, or run in reverse. The College Board files all six topics under enduring understanding FUN-3: recognizing which derivative rule applies can turn a scary-looking problem into a routine one. $$\frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) \cdot g'(x)$$ 1. 3.1 The Chain Rule: differentiate composite functions, the foundation for everything else. 2. 3.2 Implicit Differentiation: differentiate equations where $y$ is not isolated. 3. 3.3 Differentiating Inverse Functions: use the reciprocal-slope relationship $g'(x) = \frac{1}{f'(g(x))}$. 4. 3.4 Differentiating Inverse Trigonometric Functions: the $\arcsin$, $\arctan$, and $\operatorname{arcsec}$ formulas. 5. 3.5 Selecting Procedures for Calculating Derivatives: choose the right rule for any given derivative. 6. 3.6 Calculating Higher-Order Derivatives: differentiate again to reach $f''(x)$ and beyond. Topic 3.1 is the engine. A composite function is a function inside a function, like $\sin(x^2)$ or $(3x+1)^5$. The recognition cue: if evaluating it by hand means doing one thing, then doing something to the result, it is a composite and needs the chain rule. Differentiate the outer function, leave the inside untouched, then multiply by the derivative of the inside. In Leibniz form, the rule for $\frac{dy}{dx}$ is: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ Topics 3.2 through 3.4 are all the chain rule in disguise. Implicit differentiation (3.2) applies when $y$ is tangled into the equation and cannot be solved for cleanly, like $x^2 + y^2 = 25$: differentiate both sides with respect to $x$, and every $y$ term produces a $\frac{dy}{dx}$ that you solve for. For inverse functions (3.3), the graph is a reflection, so the slope $g'(x)$ is the reciprocal of the original slope at the matching point: $$g'(x) = \frac{1}{f'(g(x))}$$ Applying that to the trig functions gives the inverse trig derivatives of 3.4, such as the derivative of $\arcsin x$: $$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}$$ Topic 3.5 is where the unit's teaching pays off: given any derivative, name the rule before you compute. Is it a product, a quotient, a composite, or a combination of those? This is a decision skill, not a formula. Topic 3.6 adds higher-order derivatives, differentiating a second time to get $f''(x)$ or $\frac{d^2y}{dx^2}$ (and $f^{(n)}(x)$ in general), which Unit 5 later uses for concavity and the second derivative test. NOTE (What the exam asks): Unit 3 is 5-10% of both the AB and BC exams, but the chain rule shows up far more often than that because it is buried inside Unit 4 related rates, Unit 6 $u$-substitution, and any product or quotient with a composite inside. Expect multiple-choice items that ask you to differentiate a nested function quickly, and free-response parts that use implicit differentiation to find a tangent-line slope $\frac{dy}{dx}$ at a point. ### How to study this unit - For Topic 3.1, name the outer and inner function out loud before differentiating: for $\sqrt{4x^2+1}$ the outer is the square root and the inner is $4x^2+1$. Getting that split right is most of the chain rule. - In Topic 3.2, remember that differentiating a $y$ term always attaches $\frac{dy}{dx}$ (chain rule) while differentiating an $x$ term does not. After differentiating, collect every $\frac{dy}{dx}$ on one side and factor it out. - Do not memorize all six inverse trig derivatives in Topic 3.4 cold. Learn the three ($\arcsin$, $\arctan$, $\operatorname{arcsec}$); the cofunction versions ($\arccos$, $\operatorname{arccot}$, $\operatorname{arccsc}$) are just their negatives. On the AP exam, $\arcsin$, $\arccos$, and $\arctan$ are the ones that actually show up, so prioritize those. - Topic 3.5 is a skill, not a formula, so drill mixed sets where you cannot predict the rule. Before computing each one, label it product, quotient, chain, or a combination, then execute. - For Topic 3.6, mind your notation: the second derivative is $f''(x)$ or $\frac{d^2y}{dx^2}$. On implicit problems you usually must substitute $\frac{dy}{dx}$ back in before the second derivative simplifies. ## Unit 4: Contextual Applications of Differentiation URL: https://www.calclearn.org/units/unit-4-contextual-applications Last updated: 2026-07-22 Answer first: Unit 4 uses derivatives in real situations: reading a rate of change with correct units, analyzing straight-line motion, solving related rates, approximating with tangent lines, and evaluating 0/0 and infinity/infinity limits with L'Hospital's Rule. It is 10-15% of the AB exam and 5-10% of BC. Exam weighting: AB 10-15%, BC 5-10%. Topics: - 4.1 Interpreting the Meaning of the Derivative in Context - 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration - 4.3 Rates of Change in Applied Contexts Other Than Motion - 4.4 Introduction to Related Rates - 4.5 Solving Related Rates Problems - 4.6 Approximating Values of a Function Using Local Linearity and Linearization - 4.7 Using L'Hospital's Rule for Determining Limits of Indeterminate Forms NOTE (The core skill): Unit 4 is where the derivative stops being a rule to memorize and becomes a tool for context. It rests entirely on the differentiation techniques from Units 2 and 3, but now each topic answers a different real-world question. The skill that matters is recognition: decide what a rate means, or which setup a problem calls for, before you compute anything. Topic 4.1 fixes how you read any derivative in context: $f'(x)$ carries the units of $f$ divided by the units of $x$. If $V(t)$ is volume in liters and $t$ is in minutes, then $V'(t)$ is measured in liters per minute. Topic 4.3 extends this to applied rates that are not motion, such as how fast a tank drains or a population grows. When a problem hands you a context, read the units first: they tell you whether you need the function itself, its rate of change, or an accumulation. Topic 4.2 connects position, velocity, and acceleration for an object moving along a line. Each is the derivative of the one before it, and speed is the absolute value of velocity. The object speeds up when velocity and acceleration have the same sign and slows down when their signs differ. | Quantity | How you get it | Notation | | --- | --- | --- | | Position | Given directly | $s(t)$ | | Velocity | Derivative of position | $v(t) = s'(t)$ | | Acceleration | Derivative of velocity | $a(t) = v'(t) = s''(t)$ | | Speed | Absolute value of velocity | $|v(t)|$ | Related rates is the marquee application of the unit and a frequent free-response setup. The signal to watch for: a problem gives one rate of change and asks for another, with the two quantities tied together by a geometric or algebraic relationship. The engine is the chain rule (Topic 4.4 builds related rates on it). Write the equation relating the quantities, differentiate every variable with respect to time using $\frac{d}{dt}$, then substitute the known values. The product and quotient rules often appear when you differentiate. The habit that saves points: substitute numbers only after differentiating, never before. Topic 4.6 uses the tangent line as a stand-in for the function near the point of tangency, since a differentiable curve looks linear up close (local linearity). The linearization at $x = a$ is $L(x) = f(a) + f'(a)(x - a)$, and you evaluate it at a nearby input to approximate $f(x)$. Whether the estimate runs high or low follows from concavity: the tangent line sits below a function that is concave up (an underestimate) and above one that is concave down (an overestimate). Topic 4.7 evaluates limits that direct substitution turns into an indeterminate form. When $\lim_{x \to a} \frac{f(x)}{g(x)}$ gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$, L'Hospital's Rule says the limit equals $\lim_{x \to a} \frac{f'(x)}{g'(x)}$, differentiating the top and bottom separately. The recognition step is the whole game: confirm the form is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ first, because applying the rule to a limit that is not indeterminate can give a wrong answer. Remember that $\frac{0}{0}$ is a label for an indeterminate form, not a number, and the exam does not test other forms such as $\infty - \infty$. NOTE (What the exam asks): Unit 4 carries 10-15% of the AB exam and 5-10% of BC. Straight-line motion and related rates are staple free-response setups, and interpreting a derivative in context with correct units appears throughout both sections. Progress Check 4 mirrors this with roughly 15 multiple-choice questions and 3 free-response questions. ### How to study this unit - Topic 4.1: attach units to every rate you report. The units of $f'(x)$ are the units of $f$ per unit of $x$, so a volume rate reads as liters per minute. On free response, a correct number with missing or wrong units still loses the point. - Topic 4.2: keep velocity and speed separate. An object is speeding up only when velocity and acceleration share a sign, so check both signs instead of assuming a positive acceleration means speeding up. - Topics 4.4 and 4.5: write the relating equation, differentiate with respect to time, and only then substitute the given values. Plugging numbers in before differentiating is the most common related-rates mistake. - Topic 4.6: decide over- or underestimate from concavity. A tangent line underestimates a concave-up function and overestimates a concave-down one, so check the sign of $f''$ near the point. - Topic 4.7: verify the form is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before using L'Hospital's Rule, then differentiate numerator and denominator separately, not with the quotient rule. ## Unit 5: Analytical Applications of Differentiation URL: https://www.calclearn.org/units/unit-5-analytical-applications Last updated: 2026-07-22 Answer first: Unit 5 uses the first derivative to find where a function increases, decreases, and reaches extrema, and the second derivative for concavity and inflection points. It is the highest-weighted differentiation unit on the AB exam (15-18%), and the core skill is picking the right test: First, Second, or Candidates. Exam weighting: AB 15-20%, BC 10-15%. Topics: - 5.1 Using the Mean Value Theorem - 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points - 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing - 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema - 5.5 Using the Candidates Test to Determine Absolute (Global) Extrema - 5.6 Determining Concavity of Functions over Their Domains - 5.7 Using the Second Derivative Test to Determine Extrema - 5.8 Sketching Graphs of Functions and Their Derivatives - 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative - 5.10 Introduction to Optimization Problems - 5.11 Solving Optimization Problems - 5.12 Exploring Behaviors of Implicit Relations The whole unit rests on one move: the sign of a derivative is information about the function one level below it. Where $f'(x) > 0$ the graph of $f$ rises, where $f'(x) < 0$ it falls, and local extrema hide at critical points where $f'(x) = 0$ or $f'$ is undefined (Topic 5.2). One level up, where $f''(x) > 0$ the graph is concave up, where $f''(x) < 0$ it is concave down, and concavity flips at points of inflection (Topic 5.6). Every technique in Unit 5 is a way of reading those signs. Topics 5.1 and 5.2 open the unit with two guarantees. The Extreme Value Theorem says a function continuous on a closed interval $[a, b]$ must attain both a maximum and a minimum somewhere on it. The Mean Value Theorem says a function continuous on $[a, b]$ and differentiable on $(a, b)$ must, at some point $c$ inside, have an instantaneous rate of change equal to its average rate of change across the interval. These theorems promise that a point exists without telling you where, which is exactly what a justification-style exam question tests. $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ | Test | Reach for it when | What it tells you | | --- | --- | --- | | First Derivative Test (5.4) | You know the sign of $f'$ on both sides of a critical point | Local max if $f'$ changes from positive to negative; local min if $f'$ changes from negative to positive; neither if no sign change | | Second Derivative Test (5.7) | $f''$ is quick to evaluate at the critical point and is nonzero there | Local max if $f'' < 0$; local min if $f'' > 0$; no conclusion if $f'' = 0$ | | Candidates Test (5.5) | You need the absolute max or min on a closed interval $[a, b]$ | Compare the value of $f$ at every critical point and both endpoints; largest and smallest win | - $f'$ positive means $f$ is increasing; $f'$ negative means $f$ is decreasing (Topic 5.3). - $f'$ increasing means $f$ is concave up; $f'$ decreasing means $f$ is concave down. This is the trap: $f'$ positive and $f'$ increasing say two different things. - A sign change in $f'$ locates a local extremum of $f$; a sign change in $f''$ locates a point of inflection of $f$ (Topics 5.4, 5.6). - Given only a graph of $f'$, you can describe the shape features of $f$: its extrema, its concavity, and its inflection points (Topic 5.9). Topics 5.10 and 5.11 apply the extrema machinery to word problems: translate the scenario into a function of one variable, restrict the domain to what the context allows, then find the absolute max or min, usually with the Candidates Test. Topic 5.12 extends the same critical-point analysis to implicitly defined relations, where you locate points with $\frac{dy}{dx} = 0$ or undefined and may need $\frac{d^2y}{dx^2}$ written in terms of $x$, $y$, and $\frac{dy}{dx}$. NOTE (What the exam asks): Unit 5 carries 15-18% of the AB multiple-choice section and 8-11% of BC, making it the heaviest of the differentiation units on AB. Free-response questions here are graded on justification, not just the answer: you must name the reason, such as "$f'$ changes from positive to negative at $x = c$, so $f$ has a local maximum there." A correct extremum with no sign-based reasoning earns the point for the value but loses the justification point. ### How to study this unit - When an FRQ asks you to justify a local extremum, state the test and the sign change in words: "$f'$ changes from positive to negative at $x = c$" (Topic 5.4). A bare answer with no sign reasoning earns no justification credit. - Reach for the Candidates Test (Topic 5.5) the instant you see "absolute" or "on the closed interval $[a, b]$." Evaluate $f$, not $f'$, at every critical point and both endpoints. The endpoints are the step students forget. - Use the Second Derivative Test (Topic 5.7) only when $f''$ is fast to compute and nonzero at the critical point. If $f'' = 0$ the test is silent, so fall back to the First Derivative Test. - Drill reading a graph of $f'$ to describe $f$ (Topics 5.8-5.9). Where $f'$ is positive $f$ rises; where $f'$ is increasing $f$ is concave up. Confusing "$f'$ positive" with "$f'$ increasing" is the single most common error. - Check hypotheses before invoking an existence theorem (Topics 5.1-5.2): the Mean Value Theorem needs continuity on $[a, b]$ and differentiability on $(a, b)$, while the Extreme Value Theorem needs only continuity on $[a, b]$. ## Unit 6: Integration and Accumulation of Change URL: https://www.calclearn.org/units/unit-6-integration-and-accumulation Last updated: 2026-07-22 Answer first: Unit 6 treats integration as accumulation: adding a rate over an interval gives total change. You learn Riemann sums, the Fundamental Theorem of Calculus linking integrals to antiderivatives, and how to choose an antidifferentiation technique. At 15-20% of both AB and BC, it is a top-weighted unit. Exam weighting: AB 15-20%, BC 15-20%. Topics: - 6.1 Exploring Accumulations of Change - 6.2 Approximating Areas with Riemann Sums - 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation - 6.4 The Fundamental Theorem of Calculus and Accumulation Functions - 6.5 Interpreting the Behavior of Accumulation Functions Involving Area - 6.6 Applying Properties of Definite Integrals - 6.7 The Fundamental Theorem of Calculus and Definite Integrals - 6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation - 6.9 Integrating Using Substitution - 6.10 Integrating Functions Using Long Division and Completing the Square - 6.11 Integrating Using Integration by Parts (BC only) - 6.12 Integrating Using Linear Partial Fractions (BC only) - 6.13 Evaluating Improper Integrals (BC only) - 6.14 Selecting Techniques for Antidifferentiation The single idea beneath all of Unit 6: a definite integral adds up a rate of change to give total change. If $f$ is a rate, then $\int_a^b f(x)\,dx$ is the net accumulation from $a$ to $b$, and geometrically it is the signed area between the graph and the $x$-axis (Topic 6.1). Every other topic is a way to compute or interpret that one quantity. The topics stack in three layers. First you approximate the area with Riemann sums (left, right, midpoint) and the trapezoidal sum, and decide whether each is an over- or underestimate (Topics 6.2 and 6.3). Then the Fundamental Theorem of Calculus turns that limit of sums into an exact value (Topics 6.4 through 6.7). Finally you build the toolkit of antiderivatives that makes the FTC usable (Topics 6.8 through 6.14). NOTE (The Fundamental Theorem of Calculus, both halves): Part 1 says differentiation undoes integration: for a continuous function $f$, the accumulation function $g(x)=\int_a^x f(t)\,dt$ has derivative $g'(x)=f(x)$ (Topics 6.4 and 6.5). Part 2 says an antiderivative evaluates a definite integral: if $f$ is continuous and $F'=f$, then $\int_a^b f(x)\,dx=F(b)-F(a)$ (Topic 6.7). Part 1 tells you what an integral builds; Part 2 tells you how to compute one. $$g'(x) = f(x) \qquad \int_a^b f(x)\,dx = F(b)-F(a)$$ Method selection is the whole game in the back half of the unit, and Topic 6.14 exists only to practice it. Nobody hands you a labeled cue that says use substitution here, so the real skill is reading the integrand and matching it to a technique before you write anything down. - Basic rule or algebraic rewrite (Topics 6.8 and 6.10): the integrand is already a power, exponential, or trig form, or becomes one after long division or completing the square. Always try this first. - $u$-substitution (Topic 6.9): the integrand holds an inner function together with a constant multiple of its derivative, like $\int 2x\cos(x^2)\,dx$. Let $u$ be the inner function, and for a definite integral change the limits to $u$-values. - Integration by parts (Topic 6.11, BC only): a product of two unlike functions, such as $x e^x$ or $x\ln x$, where one factor gets simpler when differentiated. - Partial fractions (Topic 6.12, BC only): a rational function whose denominator factors into distinct linear pieces. On the exam Unit 6 carries 15-20% of the score on both AB and BC, tied for the heaviest weight. Multiple-choice questions pull Riemann sums from a table (Topic 6.2), differentiate an accumulation function (Topic 6.4), and evaluate definite integrals by the FTC (Topic 6.7). Free-response frequently gives a rate as a table or graph and asks for accumulated change with correct units (Topic 6.1). BC students also see integration by parts, partial fractions, and improper integrals (Topics 6.11 through 6.13). ### How to study this unit - Before integrating anything, run the Topic 6.14 checklist: can I use a basic rule, is there a $u$-substitution (an inner function and its derivative), or does it need integration by parts or partial fractions (BC)? Naming the method first prevents wasted work. - Lock down the over/underestimate logic for Riemann sums (Topic 6.2) by sketching: for an increasing function a left sum underestimates and a right sum overestimates, and a trapezoidal sum overestimates when the graph is concave up. - Drill the derivative-of-an-integral pattern from Topic 6.4 until it is automatic, including the chain rule when the upper limit is a function: $\frac{d}{dx}\int_a^{x^2} f(t)\,dt = f(x^2)\cdot 2x$. - On a definite-integral $u$-substitution (Topic 6.9), convert the limits to $u$-values instead of back-substituting to $x$; it is faster and kills a common sign error. - Attach units to every accumulation answer (Topic 6.1): a rate in liters per minute integrated over minutes gives liters, and FRQ graders dock points for missing or wrong units. ## Unit 7: Differential Equations URL: https://www.calclearn.org/units/unit-7-differential-equations Last updated: 2026-07-22 Answer first: Unit 7 covers differential equations, which relate a function to its derivatives. You model situations, verify solutions, read slope fields, and use separation of variables to find general and particular solutions, including exponential and logistic growth. It is 6-12% of the AB exam and 6-9% of the BC exam. Exam weighting: AB 5-10%, BC 5-10%. Topics: - 7.1 Modeling Situations with Differential Equations - 7.2 Verifying Solutions for Differential Equations - 7.3 Sketching Slope Fields - 7.4 Reasoning Using Slope Fields - 7.5 Approximating Solutions Using Euler's Method (BC only) - 7.6 Finding General Solutions Using Separation of Variables - 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables - 7.8 Exponential Models with Differential Equations - 7.9 Logistic Models with Differential Equations (BC only) NOTE (The core idea): A differential equation gives you a function's rate of change and asks you to recover the function itself. Where earlier units differentiate a known function, Unit 7 runs the process backward: you are handed $\frac{dy}{dx}$ in terms of $x$ and $y$, and your job is to find every $y$ that fits. The whole unit is one recognition question asked in different forms: given this differential equation, do I verify a solution, describe it qualitatively with a slope field, or solve it outright by separation of variables? The unit moves from setup to solution. Topic 7.1 translates a sentence into a differential equation, and Topic 7.2 checks a candidate solution by substituting it back. Topics 7.3 and 7.4 read solutions qualitatively from a slope field, a grid of short segments whose slopes are $\frac{dy}{dx}$ at each point, and BC adds Euler's method (Topic 7.5) to step along a solution numerically. Topics 7.6 and 7.7 solve equations exactly by separation of variables, Topic 7.8 applies that machinery to exponential growth, and Topic 7.9 (BC only) interprets logistic growth models qualitatively, reading carrying capacity and fastest-growth point without solving the equation. - Verify a solution (Topic 7.2): the problem hands you both an equation and a candidate function. Differentiate the candidate, substitute into the equation, and check that both sides match. You never solve, you confirm. - Slope field (Topics 7.3 and 7.4): the problem shows a grid of segments or asks you to sketch one. Plug points into $\frac{dy}{dx}$ to get slopes, then trace a solution curve by following the segments. No algebra required. - Separation of variables (Topics 7.6 and 7.7): the problem says solve or find the particular solution, and $\frac{dy}{dx}$ factors into a function of $x$ times a function of $y$. Separate the variables, integrate both sides, then apply the initial condition. $$\frac{dy}{dx} = g(x)h(y) \;\Longrightarrow\; \int \frac{1}{h(y)}\,dy = \int g(x)\,dx$$ Integrating both sides introduces a constant $+C$, so separation of variables produces a general solution: a whole family of curves, one for each value of $C$. A single initial condition, such as $y(0) = 2$, pins down one particular solution. Solve for $C$ using the initial condition right after you integrate, before simplifying, and put the $+C$ on only one side. Watch for domain restrictions: a particular solution is valid only on the interval around the initial point where it stays defined. Two word-problem patterns collapse into standard differential equations. When a rate is proportional to the current amount, the model is exponential (Topic 7.8): growth for $k > 0$, decay for $k < 0$. When the rate is jointly proportional to the amount and the room left before a carrying capacity, the model is logistic (Topic 7.9, BC only). For logistic growth you can read the carrying capacity $a$ as the value $y$ approaches as $t \to \infty$, and the population grows fastest at half the carrying capacity, both without solving the equation. $$\frac{dy}{dt} = ky \;\Longrightarrow\; y = y_0 e^{kt}$$ $$\frac{dy}{dt} = ky(a - y)$$ NOTE (What the exam asks): Unit 7 is 6-12% of the AB exam and 6-9% of the BC exam. Slope fields and separation of variables are staple free-response tasks, often paired: sketch the slope field, then solve the equation and match the two. Multiple-choice questions test verifying a solution (Topic 7.2) and setting up the right model (Topic 7.1). BC students additionally see Euler's method (Topic 7.5) and logistic models (Topic 7.9). Progress Check 7 has about 15 multiple-choice questions on AB and 20 on BC, plus 3 free-response questions. ### How to study this unit - Topics 7.2 and 7.6: read the verb. Verify a solution means substitute the given function and check that both sides agree; solve or find the general solution means separate variables and integrate. Doing the wrong one burns time on a task that wanted the other. - Topics 7.6 and 7.7: put all $y$ terms with $dy$ on one side and all $x$ terms with $dx$ on the other before integrating, write the $+C$ on the $x$ side only, and solve for $C$ from the initial condition immediately after integrating, not at the end. - Topics 7.3 and 7.4: to sketch a slope field, plug each grid point into $\frac{dy}{dx}$ and draw a short segment with that slope. When $\frac{dy}{dx}$ depends only on $x$ the segments match along vertical lines; when it depends only on $y$ they match along horizontal lines. - Topic 7.8: memorize that a rate proportional to the amount, $\frac{dy}{dt} = ky$, always solves to $y = y_0 e^{kt}$. Spotting that phrasing lets you write the answer without re-deriving it, with $k > 0$ for growth and $k < 0$ for decay. - BC only: for Euler's method (Topic 7.5) build a table and repeat $y_{i+1} = y_i + \frac{dy}{dx}\cdot \Delta x$ one step at a time; for a logistic model (Topic 7.9) read the carrying capacity as the limiting value and recall growth is fastest at half of it, both without solving. ## Unit 8: Applications of Integration URL: https://www.calclearn.org/units/unit-8-applications-of-integration Last updated: 2026-07-22 Answer first: Unit 8 turns the definite integral into a tool: integrate a rate to get accumulated change, or add up thin slices to get area and volume. It covers average value, motion, accumulation, area between curves, volumes by cross section, and solids of revolution, and it powers 10-15% of the AB exam. Exam weighting: AB 10-15%, BC 5-10%. Topics: - 8.1 Finding the Average Value of a Function on an Interval - 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals - 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts - 8.4 Finding the Area Between Curves Expressed as Functions of x - 8.5 Finding the Area Between Curves Expressed as Functions of y - 8.6 Finding the Area Between Curves That Intersect at More Than Two Points - 8.7 Volumes with Cross Sections: Squares and Rectangles - 8.8 Volumes with Cross Sections: Triangles and Semicircles - 8.9 Volume with Disc Method: Revolving Around the x- or y-Axis - 8.10 Volume with Disc Method: Revolving Around Other Axes - 8.11 Volume with Washer Method: Revolving Around the x- or y-Axis - 8.12 Volume with Washer Method: Revolving Around Other Axes - 8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled (BC only) NOTE (The one move behind the whole unit): Every problem in Unit 8 is the same move: cut the quantity into thin pieces, write down the area, volume, or change of a single piece, then integrate to add them up. The skill the exam actually rewards is method selection, looking at a problem and knowing what one slice looks like before you write a single integral. The topics fall into three strands. Topics 8.1 to 8.3 accumulate a rate: average value, motion, and accumulation functions each integrate something that changes over an interval. Topics 8.4 to 8.12 accumulate geometry: area between curves, then volumes built from cross sections, then volumes of revolution by the disc and washer methods. Topic 8.13, arc length, accumulates length and is BC only. Every strand is the limit-of-a-Riemann-sum idea from Unit 6 applied to a new quantity. $$\begin{aligned} \text{average value} &= \frac{1}{b-a}\int_a^b f(x)\,dx \\[4pt] \text{displacement} &= \int_a^b v(t)\,dt,\quad \text{distance}=\int_a^b |v(t)|\,dt \\[4pt] \text{area between curves} &= \int_a^b \big(\text{top}-\text{bottom}\big)\,dx \\[4pt] \text{volume by cross sections} &= \int_a^b A(x)\,dx \\[4pt] \text{disc, then washer} &= \pi\int_a^b R^2\,dx,\quad \pi\int_a^b (R^2-r^2)\,dx \end{aligned}$$ | The problem gives you | Recognize it as | Topics | | --- | --- | --- | | A function's average value over an interval | Average value $=\frac{1}{b-a}\int_a^b f(x)\,dx$ | 8.1 | | A rate of change and asks for a total or net amount | Net change $=\int_a^b r(t)\,dt$; total $= f(a)+\int_a^b r(t)\,dt$ | 8.3 | | Velocity, and asks how far the particle actually travels | Total distance $= \int_a^b |v(t)|\,dt$ | 8.2 | | Two curves bounding a region | Integrate top minus bottom in $x$, or right minus left in $y$ | 8.4 to 8.6 | | A base region with a named cross-section shape | Integrate that shape's area $A(x)$ along the base | 8.7, 8.8 | | A region revolved around a line, touching the axis | Disc: $\pi\int_a^b R^2\,dx$ | 8.9, 8.10 | | A region revolved with a gap to the axis | Washer: $\pi\int_a^b (R^2-r^2)\,dx$ | 8.11, 8.12 | Topic 8.1 asks for average value, which students mix up with average rate of change: average value divides an integral by the width of the interval, while average rate of change divides a change in $f$ by a change in $x$. Topic 8.2 hides the exam's favorite trap: the integral of velocity $v(t)$ gives displacement, but total distance needs the integral of speed $|v(t)|$, so you find where $v(t)$ changes sign and split the interval. Topic 8.3 problems hand you a rate and ask for a total: the net change is $\int_a^b r(t)\,dt$, and the total amount is the starting value plus that integral. For area (8.4 to 8.6), sketch the region and choose between integrating in $x$ or in $y$ by which one gives a clean top minus bottom (or right minus left) without carving the region into extra pieces; when curves cross at more than two points (8.6), split at each crossing or integrate the absolute value of the difference. For volume, first separate cross-section solids (8.7, 8.8), where you integrate the area of a named shape sitting on a base, from solids of revolution (8.9 to 8.12). Within revolution, run the gap test: if the region touches the axis, use a disc; if a gap sits between the region and the axis, use a washer. When the axis is a line other than the x- or y-axis (8.10, 8.12), write each radius as the distance from the curve to that line, not just the function value. Unit 8 is 10 to 15% of the AB exam and 6 to 9% of BC, and it pulls more than its weight on the free-response section. Expect a calculator-active accumulation question where you set up an integral exactly and evaluate it numerically, plus an area-and-volume question that usually revolves a region around some line. Multiple-choice items lean on average value, total distance, and volume setups. On the non-calculator section you antidifferentiate by hand, so the setup and the integration both earn points. ### How to study this unit - Topic 8.2 is the classic trap: read whether the question wants displacement or total distance. Displacement integrates $v(t)$; total distance integrates $|v(t)|$, which means finding where $v(t)$ changes sign and splitting the integral there. - For area (8.4, 8.5), always sketch first, then pick $dx$ or $dy$ by whichever gives one clean top-minus-bottom integral. If the curves cross more than twice (8.6), split at every intersection or integrate the absolute value of the difference. - For solids of revolution (8.9 to 8.12), run the gap test: a region touching the axis means disc, a gap means washer. When you revolve around a line that is not the x- or y-axis (8.10, 8.12), rebuild each radius as the distance from the curve to that line before squaring. - Treat Topic 8.3 accumulation problems as calculator-active: set the integral up exactly and evaluate it numerically. The net change is the integral of the rate; the total amount is the starting value plus that integral, so do not forget to add the initial amount. - Keep average value (8.1) separate from average rate of change: average value divides the integral by $(b-a)$, average rate of change divides the change in $f$ by the change in $x$. BC students, note that arc length (8.13) is the only BC-only topic in this unit. ## Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions URL: https://www.calclearn.org/units/unit-9-parametric-polar-vector Last updated: 2026-07-22 Answer first: Unit 9 is BC-only and extends calculus to curves not written as y = f(x): parametric and vector-valued functions and polar coordinates. You find tangent slopes and second derivatives, arc length, planar motion (speed, displacement, distance), and areas of polar regions. It is 10-15% of the BC exam. Exam weighting: AB n/a, BC 10-15%. Topics: - 9.1 Defining and Differentiating Parametric Equations (BC only) - 9.2 Second Derivatives of Parametric Equations (BC only) - 9.3 Finding Arc Lengths of Curves Given by Parametric Equations (BC only) - 9.4 Defining and Differentiating Vector-Valued Functions (BC only) - 9.5 Integrating Vector-Valued Functions (BC only) - 9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions (BC only) - 9.7 Defining Polar Coordinates and Differentiating in Polar Form (BC only) - 9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve (BC only) - 9.9 Finding the Area of the Region Bounded by Two Polar Curves (BC only) NOTE (The core idea): Every earlier unit worked with curves written as $y = f(x)$. Unit 9 (BC only) drops that restriction: a curve can be traced by a parameter, $x = x(t)$ and $y = y(t)$, packaged as a vector-valued function, or described by a distance and angle in polar form, $r = f(\theta)$. The calculus itself does not change. The chain rule still produces slopes and the definite integral still produces lengths and areas; you just apply them through the new variable. Every problem in the unit is one recognition question: which representation am I in, and which quantity (slope, length, motion, or area) does the problem want? The unit runs in two halves. Topics 9.1 through 9.3 are parametric calculus: the tangent slope $\frac{dy}{dx}$ (9.1), the second derivative for concavity (9.2), and arc length (9.3). Topics 9.4 through 9.6 recast the same curve as a vector-valued function to handle motion in the plane, differentiating for velocity and acceleration (9.4), integrating to recover velocity and position (9.5), and combining both for speed, displacement, and total distance (9.6). Topics 9.7 through 9.9 switch to polar coordinates: differentiating a polar curve (9.7), then the area of a single polar region (9.8) and the area between two polar curves (9.9). - Tangent slope, parametric or polar (Topics 9.1, 9.7): divide, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ when $dx/dt \neq 0$. A horizontal tangent needs $dy/dt = 0$; a vertical tangent needs $dx/dt = 0$. In polar, first convert with $x = r\cos\theta$ and $y = r\sin\theta$, then differentiate against $\theta$. - Second derivative or concavity (Topic 9.2): differentiate $\frac{dy}{dx}$ with respect to $t$, then divide by $\frac{dx}{dt}$ again. Do not divide the two second derivatives. - Arc length or total distance (Topics 9.3, 9.6): integrate speed, $\sqrt{(dx/dt)^2 + (dy/dt)^2}$, over the $t$-interval. A parametric curve's length and a particle's distance traveled use the same integrand. - Displacement versus position (Topics 9.5, 9.6): integrate the velocity vector component by component to get displacement, then add the starting position to get the final position. - Polar area (Topics 9.8, 9.9): use $\frac{1}{2}\int r^2 \,d\theta$ for one curve and $\frac{1}{2}\int (r_{\text{outer}}^2 - r_{\text{inner}}^2) \,d\theta$ between two. $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \quad (dx/dt \neq 0), \qquad \frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{dx/dt}$$ A vector-valued function bundles the two parametric equations into one position $\left\langle x(t), y(t) \right\rangle$. Differentiating component by component gives the velocity vector $\left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle$ and differentiating again gives acceleration. Speed is the magnitude of velocity, $\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$, a scalar, not a vector. Integration runs the process backward: integrate acceleration to recover velocity and velocity to recover position, using an initial condition to pin the constant on each component. Two integrals answer the classic motion questions: the definite integral of the velocity vector is displacement (net change in position, which can be zero even after a long trip), while the definite integral of speed is total distance traveled (never negative, zero only if the particle never moves). A polar curve $r = f(\theta)$ gives distance $r$ from the origin as a function of the angle $\theta$. To differentiate it, convert to $x = r\cos\theta = f(\theta)\cos\theta$ and $y = r\sin\theta = f(\theta)\sin\theta$, then apply the parametric slope rule with $\theta$ as the parameter. Area works differently from rectangular area: instead of thin rectangles you sum thin circular sectors, which is where the factor of $\frac{1}{2}$ and the $r^2$ come from. The whole challenge in Topics 9.8 and 9.9 is the limits of integration, not the formula. Find them by setting $r = 0$ (where a petal starts and ends) or, for two curves, by setting the two $r$ values equal to locate intersection angles. Sketch the region first so you know which curve is outer. $$A = \frac{1}{2}\int_{\alpha}^{\beta} \left[f(\theta)\right]^2 \,d\theta, \qquad A = \frac{1}{2}\int_{\alpha}^{\beta} \left(r_{\text{outer}}^2 - r_{\text{inner}}^2\right) \,d\theta$$ NOTE (What the exam asks): Unit 9 is BC only and worth 10-15% of the BC exam, so it carries real weight. Expect a full free-response question built on one representation: a particle in planar motion asking for speed, displacement, and total distance (Topic 9.6), or a polar region asking for area between two curves (Topic 9.9), both usually in the calculator-allowed section where you set up the integral and evaluate numerically. Multiple-choice items test the parametric slope and second derivative (Topics 9.1 and 9.2) and arc length (Topic 9.3). The recurring trap across the unit is notation: dividing the wrong pair of derivatives for $\frac{d^2y}{dx^2}$, or confusing displacement with distance. Progress Check 9 has about 25 multiple-choice questions and 3 free-response questions. ### How to study this unit - Topic 9.2: the parametric second derivative is $\frac{d^2y}{dx^2} = \frac{d/dt\left(dy/dx\right)}{dx/dt}$. Never compute $\frac{d^2y/dt^2}{d^2x/dt^2}$. Find $\frac{dy}{dx}$ first, differentiate that whole expression with respect to $t$, then divide by $\frac{dx}{dt}$ one more time. - Topics 9.3 and 9.6: arc length of a parametric curve and total distance traveled by a particle are the same integral, $\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$. Learn this integrand once and recognize both prompts as asking for it. - Topic 9.6: keep four quantities separate. Velocity is a vector, speed is its nonnegative magnitude, displacement is the integral of the velocity vector (a vector whose components can each be negative, and whose net change can be zero even after a long trip), and total distance is the integral of speed (never negative, zero only if the particle never moves). Final position equals initial position plus displacement. - Topics 9.8 and 9.9: the polar area formula is easy; the limits are the work. Sketch the curve, then find bounds by setting $r = 0$ for a single petal or by setting the two equations equal to find intersection angles, and always integrate $r_{\text{outer}}^2$ minus $r_{\text{inner}}^2$ between two curves. - Topics 9.1 and 9.7: for horizontal and vertical tangents, test the numerator and denominator of $\frac{dy}{dx}$ separately. Horizontal tangent when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$; vertical tangent when $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$. Convert polar to $x$ and $y$ first, then apply the same test. ## Unit 10: Infinite Sequences and Series URL: https://www.calclearn.org/units/unit-10-infinite-sequences-and-series Last updated: 2026-07-22 Answer first: Unit 10, BC-only, covers infinite series: deciding whether a sum of infinitely many terms settles on a finite value, and for power series, which function it represents. You learn convergence tests, two error bounds, and Taylor and Maclaurin series. It is 17-18% of the BC exam, tied with Unit 6 for the heaviest weight. Exam weighting: AB n/a, BC 15-20%. Topics: - 10.1 Defining Convergent and Divergent Infinite Series (BC only) - 10.2 Working with Geometric Series (BC only) - 10.3 The nth Term Test for Divergence (BC only) - 10.4 Integral Test for Convergence (BC only) - 10.5 Harmonic Series and p-Series (BC only) - 10.6 Comparison Tests for Convergence (BC only) - 10.7 Alternating Series Test for Convergence (BC only) - 10.8 Ratio Test for Convergence (BC only) - 10.9 Determining Absolute or Conditional Convergence (BC only) - 10.10 Alternating Series Error Bound (BC only) - 10.11 Finding Taylor Polynomial Approximations of Functions (BC only) - 10.12 Lagrange Error Bound (BC only) - 10.13 Radius and Interval of Convergence of Power Series (BC only) - 10.14 Finding Taylor or Maclaurin Series for a Function (BC only) - 10.15 Representing Functions as Power Series (BC only) NOTE (The core idea): An infinite series adds up infinitely many terms, and the whole unit turns on one question: does that running total settle on a finite number? Topics 10.1 through 10.10 answer it for a series of constants, and Topics 10.11 through 10.15 go further, letting a series of powers of $x$ stand in for an entire function. The skill the exam rewards, and a solver cannot fake, is looking at a series and knowing which test to reach for. The unit runs in two halves. Topic 10.1 defines convergence through the sequence of partial sums: a series converges exactly when that sequence has a limit. Topics 10.2 through 10.9 build the toolkit for deciding it: geometric series, the $n$th term test, the integral test, harmonic and p-series, comparison and limit comparison, the alternating series test, and the ratio test, ending with absolute versus conditional convergence. Topic 10.10 bounds the error of an alternating partial sum. The second half (Topics 10.11 through 10.15) uses these ideas to represent functions with Taylor and Maclaurin polynomials and series, the Lagrange error bound, and the radius and interval of convergence of a power series. | Series shape | Test (topic) | Converges when | | --- | --- | --- | | Terms do not shrink to $0$ | $n$th term test (10.3) | Never: it diverges | | Constant ratio, $\sum ar^n$ | Geometric (10.2) | $|r| < 1$, sum $\frac{a}{1-r}$ | | Terms like $\frac{1}{n^p}$ | p-series (10.5) | $p > 1$ | | Looks like a series you know | Comparison, limit comparison (10.6) | It beats or matches that series | | Positive and easy to integrate | Integral test (10.4) | Its improper integral converges | | Alternating, terms decreasing to $0$ | Alternating series test (10.7) | Terms decrease to $0$ | | Factorials or $n$th powers | Ratio test (10.8) | $L < 1$ | $$\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| = L \;\Rightarrow\; \begin{cases} L < 1 & \text{converges} \\ L > 1 & \text{diverges} \\ L = 1 & \text{inconclusive} \end{cases}$$ Once a series converges, Topic 10.9 asks what kind. It converges absolutely if the series of absolute values $\sum |a_n|$ also converges, and absolute convergence guarantees the original converges. A series that converges while $\sum |a_n|$ diverges converges conditionally: the alternating harmonic series $\sum \frac{(-1)^{n+1}}{n}$ is the standard example, since it converges but $\sum \frac{1}{n}$ does not. On the exam, test $\sum |a_n|$ first, then fall back to the alternating series test if that diverges. These topics turn a function into an infinite polynomial. The Taylor polynomial centered at $x = a$ is built from the derivatives of $f$ at $a$, with the $n$th-degree coefficient $\frac{f^{(n)}(a)}{n!}$ (Topic 10.11); keep every term and you have the Taylor series, or the Maclaurin series when $a = 0$. Memorize the Maclaurin series for $e^x$, $\sin x$, $\cos x$, and $\frac{1}{1-x}$ (Topic 10.14), then build new series from them by substitution, term-by-term differentiation, or term-by-term integration (Topic 10.15). A power series converges only on an interval: the ratio test gives the radius of convergence, and you test each endpoint separately to finish the interval (Topic 10.13). NOTE (The two error bounds): Two question types ask how far a partial sum or Taylor polynomial sits from the true value, and each has its own bound. Read the series first, then pick the matching bound. The alternating series error bound (Topic 10.10) applies only to a convergent alternating series: the error is at most the size of the first term you drop, $|S - S_n| \le |a_{n+1}|$. The Lagrange error bound (Topic 10.12) works for any Taylor polynomial, capping the remainder: $$|R_n(x)| \le \frac{\max |f^{(n+1)}(z)|}{(n+1)!}\,|x - a|^{n+1}$$ NOTE (What the exam asks): Unit 10 is 17-18% of the BC exam, tied with Unit 6 for the heaviest weight, and it is tested only on BC. One line from the CED settles your study list: the $n$th term test, integral test, comparison and limit comparison tests, alternating series test, and ratio test are the only convergence tests the exam assesses (Topics 10.3-10.8), so a root test or others are not needed. Multiple-choice questions lean on test selection and geometric-series sums; free-response questions build a Taylor or Maclaurin series, find its interval of convergence, or bound an error. Progress Check 10 has about 45 multiple-choice questions and 3 free-response questions. ### How to study this unit - Build a convergence-test reflex: run the $n$th term test first (Topic 10.3), then check whether the series is geometric (Topic 10.2) or a p-series (Topic 10.5) you can classify on sight, and save the ratio test (Topic 10.8) for terms with factorials or $n$th powers. - Remember the $n$th term test (Topic 10.3) can only prove divergence. If $\lim_{n \to \infty} a_n = 0$ the series may still diverge: the harmonic series $\sum \frac{1}{n}$ (Topic 10.5) is the classic trap where terms go to $0$ but the sum grows without bound. - Memorize the four core Maclaurin series, $e^x$, $\sin x$, $\cos x$, and $\frac{1}{1-x}$ (Topic 10.14), then generate everything else by substituting, differentiating, or integrating term by term (Topic 10.15) instead of recomputing derivatives from scratch. - For an interval of convergence (Topic 10.13), use the ratio test to get the radius, then test each endpoint separately with a different test. The ratio test is always inconclusive at the endpoints, so a series can converge at one, both, or neither. - Match the error bound to the series: the alternating series error bound (Topic 10.10) is just the size of the first dropped term and only works for alternating series, while the Lagrange error bound (Topic 10.12) needs the max of the next derivative and works for any Taylor polynomial.