AP Calculus AB and BC

Derivative of sin x / x^2: Quotient Rule

The derivative of sin x over x squared is x cos x minus 2 sin x, all over x cubed. The quotient rule produces an x squared and an x term on top, and cancelling one factor of x from every term is what reduces the denominator from x to the fourth to x cubed.

ddx[sin⁡xx2]=xcos⁡x−2sin⁡xx3\frac{d}{dx}\left[\frac{\sin x}{x^{2}}\right] = \frac{x\cos x - 2\sin x}{x^{3}}

Quotient rule, then simplify

x2cos⁡x−sin⁡x⋅2xx4=x(xcos⁡x−2sin⁡x)x4=xcos⁡x−2sin⁡xx3\frac{x^{2}\cos x - \sin x\cdot 2x}{x^{4}} = \frac{x\left(x\cos x - 2\sin x\right)}{x^{4}} = \frac{x\cos x - 2\sin x}{x^{3}}

Every term on top carries a factor of xx, so cancelling it is not optional tidying: leaving x4x^{4} underneath makes the expression look more singular at the origin than it is.

Behaviour at the origin

Neither the function nor its derivative exists at x=0x = 0. Since sin⁡x≈x\sin x \approx x near zero, the function behaves like 1x\frac{1}{x} there, so it has a vertical asymptote rather than a removable hole.

Common mistakes

  • Getting the numerator order backwards. It is u′v−uv′u'v - uv'.
  • Leaving x4x^{4} in the denominator without cancelling.

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 / x^2?

It is xcos⁡x−2sin⁡xx3\frac{x\cos x - 2\sin x}{x^{3}}.

Is there a hole at x = 0?

No. Unlike sin⁡xx\frac{\sin x}{x}, this behaves like 1x\frac{1}{x} near zero, so it has a vertical asymptote.