Mathematics 9709/23 — May/June 2022

Walkthrough

The function $y = \frac{\ln x}{x^2}$ is a quotient, so the quotient rule is the natural method. Write the numerator as $u = \ln x$ and the denominator as $v = x^2$. Differentiate these separately: $\frac{du}{dx} = \frac{1}{x}$ and $\frac{dv}{dx} = 2x$. The quotient rule states that

Substituting gives $\frac{x^2 \cdot \frac{1}{x} - 2x\ln x}{x^4} = \frac{x - 2x\ln x}{x^4}$. This is the derivative in unsimplified form.

To evaluate at $x = e$, use the key fact $\ln e = 1$. Then the numerator becomes $e - 2e \cdot 1 = -e$, and the denominator is $e^4$. Therefore the value is $\frac{-e}{e^4} = -\frac{1}{e^3}$. An equivalent way is to write $y = x^{-2}\ln x$ and use the product rule, which gives $\frac{dy}{dx} = -2x^{-3}\ln x + x^{-3}$; at $x=e$ this also gives $-\frac{2}{e^3} + \frac{1}{e^3} = -\frac{1}{e^3}$.

Key Takeaways

This question tests the quotient rule (or product rule) together with the derivative of $\ln x$. It also checks exact evaluation: you must know $\ln e = 1$ and simplify powers of $e$ correctly. Recognising that $\frac{e}{e^4} = \frac{1}{e^3}$ is an important algebraic skill.

Common Mistakes

• Forgetting the minus sign in the quotient rule: the numerator must be $v\frac{du}{dx} - u\frac{dv}{dx}$, not $u\frac{dv}{dx} - v\frac{du}{dx}$.
• Differentiating $x^2$ incorrectly, or writing the denominator as $x^2$ instead of $x^4$.
• Substituting $x = e$ but forgetting that $\ln e = 1$, leaving $\ln e$ in the answer.
• Not simplifying $\frac{-e}{e^4}$ to $-\frac{1}{e^3}$.
• Giving a decimal approximation instead of the exact value.

Things to Be Careful About

• The derivative must be found using a valid method; the mark scheme awards a method mark for using the quotient or product rule, so show your working.
• When using the product rule, remember that $x^{-2}$ differentiates to $-2x^{-3}$, not $2x^{-3}$.
• At the substitution stage, keep the expression exact. Do not replace $e$ with a decimal.
• Check signs carefully: $e - 2e = -e$, not $e$.