AP Calculus AB and BC

Derivative of sin x: Answer, Proof, Mistakes

Q: Why is the derivative of sin x equal to cos x and not -cos x?

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?

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)?

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?

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.

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

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

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.
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.
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.
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.
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.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

Why is the derivative of sin x equal to cos x and not -cos x?

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 $-\sin x \cdot 1$ .

Does the formula only work in radians?

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.

What is the derivative of sin(2x)?

Apply the chain rule. The outer derivative is $\cos(2x)$ and the inner derivative of $2x$ is $2$ , so $\frac{d}{dx}[\sin(2x)] = 2\cos(2x)$ .

What is the second derivative of sin x?

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 $\cos x$ , $-\sin x$ , $-\cos x$ , $\sin x$ .