AP Calculus AB and BC

Derivative of cos x: Answer, Proof, Mistakes

Q: Is the derivative of cos x sin x or negative sin x?

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?

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?

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?

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.

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 $x > 0$ .

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.

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.

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

Is the derivative of cos x sin x or negative sin x?

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: $\frac{d}{dx}[\sin x] = \cos x$ .

Why does the derivative of cos x have a minus sign?

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 $\cos(x+h) = \cos x\cos h - \sin x\sin h$ .

What is the derivative of cos x squared?

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 $-2x\sin(x^2)$ .

What is the second derivative of cos x?

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 $y'' = -y$ .