AP Calculus AB and BC

Derivative of sin x cos x: Answer, Proof, Mistakes

Q: What is the derivative of sin x cos x?

\frac{d}{dx}(\sin x\cos x) = \cos 2x. The product rule gives \cos^{2}x - \sin^{2}x, and the double-angle identity rewrites that as \cos 2x.

Q: Can I use a double-angle identity instead of the product rule?

Yes, and it is faster. Since \sin x\cos x = \frac{1}{2}\sin 2x, one chain rule gives \frac{1}{2}\cdot 2\cos 2x = \cos 2x, the same result.

The derivative of sin x cos x with respect to x is cos 2x. The product rule gives cos x times cos x plus sin x times negative sin x, which is cos^2 x - sin^2 x, and the double-angle identity rewrites that as cos 2x. You can also differentiate the equivalent form (1/2) sin 2x to reach the same answer.

\frac{d}{dx}\left[\sin x\cos x\right] = \cos 2x

How to differentiate sin x cos x

Treat it as a product of $\sin x$ and $\cos x$ and apply the product rule, differentiating one factor at a time.

\frac{d}{dx}\left(\sin x\cos x\right) = \cos x\cdot\cos x + \sin x\cdot(-\sin x)

That simplifies to a difference of squares, which is exactly the double-angle formula for cosine.

\cos^{2}x - \sin^{2}x = \cos 2x

The faster route: use the identity first

The double-angle identity $\sin x\cos x = \frac{1}{2}\sin 2x$ lets you rewrite before differentiating. Then a single chain rule finishes it.

\frac{d}{dx}\left(\frac{1}{2}\sin 2x\right) = \frac{1}{2}\cdot 2\cos 2x = \cos 2x

Both methods land on $\cos 2x$ , which is a good check that the product rule and the identity agree.

Where the derivative of sin x cos x shows up on the AP exam

The product rule is Topic 2.8 and the identity route uses the chain rule in Topic 3.1, both on AB and BC. This function is a common place to test whether you recognize the double-angle identities, which also appear when integrating.

The reverse shows up in Unit 6: $\int \sin x\cos x\,dx = \frac{1}{2}\sin^{2}x + C$ , or equivalently $-\frac{1}{4}\cos 2x + C$ , forms that differ only by a constant.

Common mistakes with the derivative of sin x cos x

Multiplying the two derivatives to get $-\sin x\cos x$ . The product rule adds two terms, it does not multiply the derivatives.
Leaving the answer as $\cos^{2}x - \sin^{2}x$ without recognizing it equals $\cos 2x$ . Both are correct, but the compact form is what later steps expect.
Answering $2\sin x\cos x$ by confusing this with the derivative of $\sin^{2}x$ . That is a different function, equal to $\sin 2x$ .
Writing $\sin^{2}x - \cos^{2}x$ , which is $-\cos 2x$ . The $\cos x\cos x$ term is positive; the minus attaches to the $\sin x\sin x$ term.

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

What is the derivative of sin x cos x?

$\frac{d}{dx}(\sin x\cos x) = \cos 2x$ . The product rule gives $\cos^{2}x - \sin^{2}x$ , and the double-angle identity rewrites that as $\cos 2x$ .

Can I use a double-angle identity instead of the product rule?

Yes, and it is faster. Since $\sin x\cos x = \frac{1}{2}\sin 2x$ , one chain rule gives $\frac{1}{2}\cdot 2\cos 2x = \cos 2x$ , the same result.

Is cos^2 x - sin^2 x the same as cos 2x?

Yes. It is one of the double-angle identities for cosine, alongside $2\cos^{2}x - 1$ and $1 - 2\sin^{2}x$ . All three equal $\cos 2x$ , so the product-rule answer and the compact answer agree.