AP Calculus AB and BC

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

The derivative of sin x / x with respect to x is (x cos x - sin x)/x^2. The quotient rule differentiates the top to cos x and the bottom to 1, which gives x cos x minus sin x, all over x squared. This is the slope formula, not the famous limit of sin x / x as x approaches 0, which equals 1.

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

The proof: quotient rule on sin x over x

Put f(x)=sin⁡xf(x) = \sin x on top and g(x)=xg(x) = x underneath, so f′(x)=cos⁡xf'(x) = \cos x and g′(x)=1g'(x) = 1.

ddx(sin⁡xx)=cos⁡x⋅x−sin⁡x⋅1x2=xcos⁡x−sin⁡xx2\frac{d}{dx}\left(\frac{\sin x}{x}\right) = \frac{\cos x\cdot x - \sin x\cdot 1}{x^2} = \frac{x\cos x - \sin x}{x^2}

The order inside the numerator decides every sign in the answer. Derivative of the top times the bottom comes first, so sin⁡x−xcos⁡x\sin x - x\cos x would give the negative of the truth.

The formula is valid for every x≠0x \neq 0, which matches the domain of the original quotient. Topic 2.9 (The Quotient Rule) is where this sits on both exams.

The limit of sin x over x is a different fact

Most students meet sin⁡xx\frac{\sin x}{x} first as a limit rather than as a function to differentiate, and the two facts get mixed up.

lim⁡x→0sin⁡xx=1\lim_{x \to 0}\frac{\sin x}{x} = 1

That limit is the difference quotient for sin⁡x\sin x at x=0x = 0. It is proved with the Squeeze Theorem in Topic 1.8, and it is the reason ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x holds at all. It says nothing directly about the slope of sin⁡xx\frac{\sin x}{x}.

The quotient has a removable discontinuity at x=0x = 0. Patch the hole with the value 1 and the patched function is differentiable there with slope 00, which is what xcos⁡x−sin⁡xx2\frac{x\cos x - \sin x}{x^2} approaches as x→0x \to 0.

Common mistakes

  • Answering cos⁡x\cos x by differentiating the top and the bottom separately. Test it at x=πx = \pi: the correct slope is −1π-\frac{1}{\pi}, while cos⁡π=−1\cos\pi = -1.
  • Reversing the numerator to sin⁡x−xcos⁡xx2\frac{\sin x - x\cos x}{x^2}, the negative of the correct derivative.
  • Dividing by xx instead of x2x^2. The quotient rule always squares the bottom.
  • Quoting lim⁡x→0sin⁡xx=1\lim_{x \to 0}\frac{\sin x}{x} = 1 as though it were the derivative. It is the height the function approaches, not its slope.

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?

It is xcos⁡x−sin⁡xx2\frac{x\cos x - \sin x}{x^2}, by the quotient rule.

Is the derivative of sin x / x just cos x?

No. Check x=πx = \pi: the real derivative is π(−1)−0π2=−1π\frac{\pi(-1) - 0}{\pi^2} = -\frac{1}{\pi}, about −0.318-0.318, while cos⁡π=−1\cos\pi = -1. Differentiating numerator and denominator separately is never the quotient rule.

What happens to the derivative at x = 0?

The formula is undefined at 00 because of the x2x^2, but the hole in sin⁡xx\frac{\sin x}{x} is removable. Defining the value at 00 to be 1 makes the function differentiable there with slope 00.