Write the minimum value

Solution:

Given: $f(x)=x^{x}$

Taking log on both sides, we get

$\log f(x)=x \log x$

Differentiating w.r.t. $x$, we get

$\frac{1}{f(x)} f^{\prime}(x)=\log x+1$

$\Rightarrow f^{\prime}(x)=f(x)(\log x+1)$

$\Rightarrow f^{\prime}(x)=x^{x}(\log x+1)$            .......(1)

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

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

$\Rightarrow x^{x}(\log x+1)=0$

$\Rightarrow \log x=-1$

$\Rightarrow x=\frac{1}{e}$

Now,

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

At $x=\frac{1}{e}:$

$f^{\prime \prime}\left(\frac{1}{e}\right)=\frac{1}{e}^{\frac{1}{e}}\left(\log \frac{1}{e}+1\right)^{2}+\frac{1}{e}^{\frac{1}{e}-1}=\frac{1}{e}^{\frac{1}{e}-1}>0$

So, $x=\frac{1}{e}$ is a point of local minimum.

Thus, the minimum value is given by

$f\left(\frac{1}{e}\right)=\frac{1}{e}^{\frac{1}{e}}=e^{\frac{-1}{e}}$