Show that the function given by $f(x) = \cfrac{{\log x}}{x}$ has maximum at $x = e$.

We have, $f(x) = \cfrac{{\log x}}{x},x > 0$ …(i) Differentiating (i) w.r.t. $x$, we get
$f'(x) = \cfrac{{x\left( {\cfrac{1}{x}} \right) - (\log x) \cdot 1}}{{{x^2}}} = \cfrac{{1 - \log x}}{{{x^2}}}$ …(ii)

For maximum or minimum, $f'(x) = 0$
$\Rightarrow \cfrac{{1 - \log x}}{{{x^2}}} = 0 \Rightarrow \log x = 1$
$\Rightarrow x = e$

Now differentiating (ii) w.r.t. $x$, we get

$f''(x) = \cfrac{{{x^2}\left( { - \cfrac{1}{x}} \right) - (1 - \log x)2x}}{{{x^4}}} = \cfrac{{ - x - 2x + 2x\log x}}{{{x^4}}}$

$= \cfrac{{x(2\log x - 3)}}{{{x^4}}} = \cfrac{{2\log x - 3}}{{{x^3}}}$

Also, $f''(e) = \cfrac{{2\log e - 3}}{{{e^3}}} = \cfrac{{2 \cdot 1 - 3}}{{{e^3}}} = \cfrac{{ - 1}}{{{e^3}}} < 0$

$\Rightarrow f(x)$ has maximum at $x = e.$