Write the maximum value

Solution:

Given : $f(x)=\frac{\log x}{x}$

$\Rightarrow f^{\prime}(x)=\frac{1-\log x}{x^{2}}$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow \frac{1-\log x}{x^{2}}=0$

$\Rightarrow 1-\log x=0$

$\Rightarrow \log x=1$

$\Rightarrow \log x=\log e$

$\Rightarrow x=e$

Now,

$f^{\prime \prime}(x)=\frac{-x-2 x(1-\log x)}{x^{4}}=\frac{-3 x-2 x \log x}{x^{4}}$

At $x=e:$

$f^{\prime \prime}(e)=\frac{-3 e-2 e \log e}{e^{4}}=\frac{-5}{e^{3}}<0$

So, $x=e$ is a point of local maximum.

Thus, the local maximum value is given by

$f(e)=\frac{\log e}{e}=\frac{1}{e}$