differentiation again.

We have to use implicit differentiation.

$y=x^x\\ \ln y = x \ln x\\ \dfrac{\mathtt dy}{\mathtt dx}\dfrac{1}{y}=\ln x+1\text{ <-- product rule}\\ \dfrac{\mathtt dy}{\mathtt dx}=y(\ln x + 1)\\ \;\;\;\;\;\;\!=x^x\ln x + x^x$