Answer
$x^x +x^x\ln x$
Work Step by Step
We are given
$$h(x)=x^x$$
From Exercise 53 we have the idea:
$\frac{d}{dx}h(x)=u(x)^{v(x)}[\frac{v(x)u'(x)}{u(x)}+(\ln u(x)v'(x))]$
so:
$\frac{d}{dx}h(x)=x^x[\frac{x}{x}+(\ln x)]=x^x[1+\ln x]=x^x +x^x\ln x$
