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

The proof: quotient rule on x/ln x

The quotient rule multiplies the derivative of the top by the bottom, subtracts the top times the derivative of the bottom, and divides the whole thing by the bottom squared.

Here $f(x) = x$ and $g(x) = \ln x$, so $f'(x) = 1$ and $g'(x) = \frac{1}{x}$.

The subtracted term collapses because $x\cdot\frac{1}{x} = 1$, so the numerator is exactly $\ln x - 1$. Keep the denominator written as $(\ln x)^2$, the square of the logarithm, which is a different object from $\ln(x^2)$.

Domain, the asymptote at x = 1, and the minimum at e

Two conditions restrict the domain. The logarithm needs $x > 0$, and the denominator needs $\ln x \neq 0$, which throws out $x = 1$. So the function lives on $0 < x < 1$ and on $x > 1$, with a vertical asymptote separating the two pieces.

Setting the numerator to zero gives $\ln x = 1$, so $x = e$. Just left of $e$ the numerator is negative and just right of it the numerator is positive, so the First Derivative Test marks $x = e$ as a minimum of the branch $x > 1$.

On $0 < x < 1$ the numerator $\ln x - 1$ is negative while $(\ln x)^2$ is positive, so the function decreases across that entire piece. The differentiation itself is Topic 2.9 (The Quotient Rule); the sign analysis is Unit 5 work.

Common mistakes

• Differentiating top and bottom separately to get $\frac{1}{\frac{1}{x}} = x$. That is the shape of L'Hopital's Rule, which applies to limits of indeterminate forms, not to derivatives of quotients.
• Reversing the numerator to $\frac{1 - \ln x}{(\ln x)^2}$, which is the negative of the correct answer. Derivative of the top times the bottom comes first.
• Writing the denominator as $\ln(x^2)$. That equals $2\ln x$, not $(\ln x)^2$.
• Forgetting that $x = 1$ is excluded. The derivative formula fails there too, since $(\ln 1)^2 = 0$.