AP Calculus AB and BC

Derivative of x^3 sin x: Product Rule

The derivative of x cubed times sin x is 3x squared sin x plus x cubed cos x. The product rule differentiates each factor while holding the other fixed, and the answer factors as x squared times 3 sin x plus x cos x.

ddx[x3sin⁡x]=3x2sin⁡x+x3cos⁡x\frac{d}{dx}\left[x^{3}\sin x\right] = 3x^{2}\sin x + x^{3}\cos x

The product rule

ddx(x3sin⁡x)=3x2sin⁡x+x3cos⁡x=x2(3sin⁡x+xcos⁡x)\frac{d}{dx}\left(x^{3}\sin x\right) = 3x^{2}\sin x + x^{3}\cos x = x^{2}\left(3\sin x + x\cos x\right)

Factoring afterwards is worth the extra line: it shows x=0x = 0 is a critical point, and that the remaining ones solve 3sin⁡x+xcos⁡x=03\sin x + x\cos x = 0.

Reversing it needs parts three times

∫x3sin⁡x dx\int x^{3}\sin x\,dx drops the power by one on each pass, so it takes three applications of integration by parts. The derivative takes one line; the antiderivative takes a page.

Common mistakes

  • Multiplying the derivatives to get 3x2cos⁡x3x^{2}\cos x.
  • Losing the second term entirely.

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 x^3 sin x?

It is 3x2sin⁡x+x3cos⁡x3x^{2}\sin x + x^{3}\cos x.

Where are the critical points?

At x=0x = 0 and wherever 3sin⁡x+xcos⁡x=03\sin x + x\cos x = 0, which needs a numerical solve.