AP Calculus AB and BC

Derivative of (ln x)/x: Answer, Proof, Mistakes

Q: What is the derivative of \frac{\ln x}{x}?

It is \frac{1-\ln x}{x^{2}} for x>0, from the quotient rule with u = \ln x and v = x.

Q: Why is the derivative zero at x = e?

Because \ln e = 1 makes the numerator 1-\ln x vanish. That point is the maximum of \frac{\ln x}{x}, with value \frac{1}{e}.

Q: Can the product rule give the same answer?

Yes. Write the function as x^{-1}\ln x and the product rule gives -x^{-2}\ln x+x^{-2}, which is \frac{1-\ln x}{x^{2}}.

The derivative of (ln x)/x is (1 - ln x)/x^2, valid for x > 0. In prime notation, if f(x) = (ln x)/x then f'(x) = (1 - ln x)/x^2. The quotient rule gives (x times 1/x minus ln x times 1) over x^2, and the first product collapses to 1.

\frac{d}{dx}\left[\frac{\ln x}{x}\right] = \frac{1-\ln x}{x^2}

How to differentiate ln x over x

Take $u = \ln x$ on top and $v = x$ underneath, then apply the quotient rule $\frac{u'v-uv'}{v^2}$ with $u' = \frac{1}{x}$ and $v' = 1$ .

\frac{d}{dx}\left(\frac{\ln x}{x}\right) = \frac{\frac{1}{x}\cdot x-\ln x\cdot 1}{x^{2}} = \frac{1-\ln x}{x^{2}}

The product $\frac{1}{x}\cdot x$ simplifies to $1$ , which is why such a short answer falls out. The domain is $x>0$ , inherited from $\ln x$ .

Rewriting as a product gives the same result and is often faster under time pressure.

\frac{d}{dx}\left(x^{-1}\ln x\right) = -x^{-2}\ln x+x^{-1}\cdot\frac{1}{x} = \frac{1-\ln x}{x^{2}}

Why the derivative is zero at x = e

On the domain $x>0$ the denominator $x^{2}$ is positive, so the sign of $f'$ is the sign of $1-\ln x$ . That is positive for $0<x<e$ and negative for $x>e$ .

f'(e) = \frac{1-\ln e}{e^{2}} = 0, \qquad f(e) = \frac{1}{e}

So $\frac{\ln x}{x}$ climbs to a single maximum height of $\frac{1}{e}$ at $x=e$ and decreases forever after. One classic consequence: $\frac{\ln \pi}{\pi}<\frac{\ln e}{e}$ , which rearranges to $\pi^{e}<e^{\pi}$ .

Common mistakes with the derivative of (ln x)/x

Differentiating the top and the bottom separately, so the answer comes out as $\frac{1}{x}$ over $1$ , that is $\frac{1}{x}$ . The quotient rule is not term by term.
Reversing the numerator and writing $\frac{\ln x-1}{x^{2}}$ , which is the negative of the correct answer. The critical point is still $x=e$ , but every increasing and decreasing interval flips, so the maximum gets misread as a minimum.
Squaring the wrong thing: the denominator is $x^{2}$ , the square of $v = x$ , not $x$ itself.
Ignoring the domain and describing behavior for $x\le 0$ , where $\ln x$ does not exist.

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 $\frac{\ln x}{x}$ ?

It is $\frac{1-\ln x}{x^{2}}$ for $x>0$ , from the quotient rule with $u = \ln x$ and $v = x$ .

Why is the derivative zero at $x = e$ ?

Because $\ln e = 1$ makes the numerator $1-\ln x$ vanish. That point is the maximum of $\frac{\ln x}{x}$ , with value $\frac{1}{e}$ .

Can the product rule give the same answer?

Yes. Write the function as $x^{-1}\ln x$ and the product rule gives $-x^{-2}\ln x+x^{-2}$ , which is $\frac{1-\ln x}{x^{2}}$ .