If $y=x^x, \frac{d y}{d x}$ will be:

$x^x(1+\log x)$

$y=x^x$    .......(1)

taking $\log$ on both sides

$\log y=\log x^x$

$\Rightarrow \log y=x \log x$

as $\log a^b$

$=b \log a$

diffecenttating both sides w.r.t (x)

$\frac{d}{d x}(\log y)=\frac{d}{d x}(x \log x)$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=x \frac{d}{d x}(\log x)+\log x \frac{d x}{d x}$

using product rule

$\Rightarrow \frac{1}{y} \frac{d y}{d x} =\frac{x}{x}+\log x$

$\frac{d y}{d x} =y(1+\log x)$

Substituting y from eq (1)

$\frac{d y}{d x}=x^x(1+\log x)$