AP Calculus AB and BC

Derivative of sin^3 x: Answer, Proof, and Mistakes

The derivative of sin^3 x is 3 sin^2(x) cos(x). In prime notation, if f(x) = sin^3 x then f'(x) = 3 sin^2(x) cos(x). Read sin^3 x as (sin x)^3, differentiate the outer cube into 3(sin x)^2, then multiply by the inner derivative cos x.

ddx[sin⁡3x]=3sin⁡2xcos⁡x\frac{d}{dx}\left[\sin^3 x\right] = 3\sin^2 x\cos x

How to differentiate sin^3 x

The notation sin⁡3x\sin^3 x means (sin⁡x)3(\sin x)^3. The outer function is the cube and the inner is sin⁡x\sin x, so use ddxu3=3u2 u′\frac{d}{dx}u^3 = 3u^2\,u' with u=sin⁡xu = \sin x and u′=cos⁡xu' = \cos x.

ddx(sin⁡x)3=3(sin⁡x)2cos⁡x=3sin⁡2xcos⁡x\frac{d}{dx}(\sin x)^3 = 3(\sin x)^2\cos x = 3\sin^2 x\cos x

The cos⁡x\cos x at the end is the inner derivative; without it you would only have the outer power's contribution.

Reading the notation, and where it appears

sin⁡3x\sin^3 x is a power of a trig function, differentiated with the chain rule (Topic 3.1). Do not confuse it with sin⁡(x3)\sin(x^3), where the cube is inside the sine and the derivative is 3x2cos⁡(x3)3x^2\cos(x^3) instead.

Powers of sine and cosine like this are common on both AB and BC, and they reappear on BC inside trig integrals, where the same sin⁡2xcos⁡x\sin^2 x\cos x pattern signals a u=sin⁡xu = \sin x substitution.

Common mistakes with the derivative of sin^3 x

  • Answering 3sin⁡2x3\sin^2 x and dropping the inner derivative cos⁡x\cos x.
  • Reading sin⁡3x\sin^3 x as sin⁡(x3)\sin(x^3) and answering 3x2cos⁡(x3)3x^2\cos(x^3). Those are different functions.
  • Writing 3cos⁡2xcos⁡x3\cos^2 x\cos x, differentiating the inside into the outer power instead of leaving sin⁡2x\sin^2 x in place.

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⁡3x\sin^3 x?

It is 3sin⁡2xcos⁡x3\sin^2 x\cos x, from the chain rule with outer cube and inner sin⁡x\sin x.

How is sin⁡3x\sin^3 x different from sin⁡(x3)\sin(x^3)?

In sin⁡3x=(sin⁡x)3\sin^3 x = (\sin x)^3 you take the sine first and cube it, giving 3sin⁡2xcos⁡x3\sin^2 x\cos x. In sin⁡(x3)\sin(x^3) you cube first, giving 3x2cos⁡(x3)3x^2\cos(x^3).

What is the derivative of cos⁡3x\cos^3 x?

It is −3cos⁡2xsin⁡x-3\cos^2 x\sin x, the same structure but with the inner derivative −sin⁡x-\sin x supplying a minus sign.