Problem Statement
Consider the function $$f(x) = \frac{\ln x}{x^2}$$ for $$x > 0$$.\n\n(a) Find $$f'(x)$$. [3 marks]\n\n(b) Hence, find the coordinates of the local maximum point on the graph of $$f$$. [3 marks]
Verified Solution & Marking Scheme
Part (a): Quotient Rule
Using the quotient rule $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$$ with $$u = \ln x$$ ($$u' = \frac{1}{x}$$) and $$v = x^2$$ ($$v' = 2x$$):\n\n$$f'(x) = \frac{\left(\frac{1}{x}\right)(x^2) - (\ln x)(2x)}{(x^2)^2} = \frac{x - 2x\ln x}{x^4} = \frac{1 - 2\ln x}{x^3}$$
Part (b): Find stationary point
Set $$f'(x) = 0$$:\n$$\frac{1 - 2\ln x}{x^3} = 0 \implies 1 - 2\ln x = 0 \implies \ln x = \frac{1}{2} \implies x = e^{1/2} = \sqrt{e}$$
Calculate corresponding y-coordinate
$$y = f(e^{1/2}) = \frac{\ln(e^{1/2})}{(e^{1/2})^2} = \frac{1/2}{e} = \frac{1}{2e}$$
Confirm local maximum
For $$x < \sqrt{e}$$, $$f'(x) > 0$$; for $$x > \sqrt{e}$$, $$f'(x) < 0$$. Hence $$\left(\sqrt{e}, \frac{1}{2e}\right)$$ is indeed a local maximum.