AP Calculus AB and BC

Derivative of sin^2 x: Answer, Chain Rule, Mistakes

The derivative of sin squared x with respect to x is 2 sin x cos x, which the double angle identity rewrites as sin 2x. The outer function is the squaring and the inner function is sine, so the chain rule gives twice sine times the derivative of sine.

ddx[sin⁡2x]=2sin⁡xcos⁡x\frac{d}{dx}\left[\sin^{2} x\right] = 2\sin x\cos x

How to differentiate sin^2 x

The notation sin⁡2x\sin^{2} x means (sin⁡x)2\left(\sin x\right)^{2}, so the squaring happens LAST and is therefore the outer function.

ddx(sin⁡x)2=2(sin⁡x)1⋅cos⁡x\frac{d}{dx}\left(\sin x\right)^{2} = 2\left(\sin x\right)^{1}\cdot\cos x
ddxsin⁡2x=2sin⁡xcos⁡x=sin⁡2x\frac{d}{dx}\sin^{2} x = 2\sin x\cos x = \sin 2x

Both forms are correct

The double angle identity sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x means the two answers are the same function. If your answer and the key disagree, check this identity before assuming an error.

The companion result is ddxcos⁡2x=−2sin⁡xcos⁡x\frac{d}{dx}\cos^{2} x = -2\sin x\cos x, the exact negative. Adding the two gives zero, which is right, because sin⁡2x+cos⁡2x=1\sin^{2}x + \cos^{2}x = 1 is constant.

Where the derivative of sin^2 x shows up on the AP exam

Topic 3.1, on both exams, and it is one of the clearest tests of whether a student can identify which function is outer. It also arrives inside volume problems, where squaring a sine radius is routine.

Do not confuse it with sin⁡(x2)\sin\left(x^{2}\right), whose derivative is 2xcos⁡(x2)2x\cos\left(x^{2}\right). The notation differs by where the square sits, and so do the answers.

Common mistakes with the derivative of sin^2 x

  • Answering 2sin⁡x2\sin x and forgetting the inner derivative cos⁡x\cos x.
  • Answering cos⁡2x\cos^{2} x, differentiating the inside and the outside at once.
  • Reading sin⁡2x\sin^{2} x as sin⁡(x2)\sin\left(x^{2}\right). They are different functions with different derivatives.

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^2 x?

It is 2sin⁡xcos⁡x2\sin x\cos x, equivalently sin⁡2x\sin 2x.

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

Yes, by the double angle identity. Either form is a correct answer.

What is the derivative of cos^2 x?

It is −2sin⁡xcos⁡x-2\sin x\cos x, or −sin⁡2x-\sin 2x, the exact negative of this one.

How is sin^2 x different from sin(x^2)?

In sin⁡2x\sin^{2}x you take sine first and square second, giving 2sin⁡xcos⁡x2\sin x\cos x. In sin⁡(x2)\sin(x^{2}) you square first, giving 2xcos⁡(x2)2x\cos(x^{2}).