Derivatives of Trig Functions: Table and Memory Tricks

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.

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.

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.

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.

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.