The maximum value of $(\frac{1}{x})^x$ is :

$e^{\frac{1}{e}}$

The correct answer is Option (3) → $e^{\frac{1}{e}}$

$\frac{1}{x^x}$

$\log y=-x\log x$ differentiating wrt x

$\frac{1}{y}\frac{dy}{dx}=-\log x-1⇒\frac{dy}{dx}=-\frac{(\log x+1)}{x^x}$

so at optimize point

$\frac{dy}{dx}=0⇒\log x=-1$

$⇒x=\frac{1}{e}$

max. value → $y(\frac{1}{e})=e^{\frac{1}{e}}$