What is the maximum value of $f(x) = \frac{\log x}{x}$ for $x > 0$?

Explanation

1. 01

   $f'(x) = \frac{1 - \log x}{x^{2}} = 0$ at $\log x = 1$, that is $x = e$.

2. 02

   $f'$ is positive before $e$ and negative after, so the maximum value is $f(e) = \frac{1}{e}$.