## Calculus with Applications (10th Edition)

$x^x +x^x\ln x$
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: $\frac{d}{dx}h(x)=x^x[\frac{x}{x}+(\ln x)]=x^x[1+\ln x]=x^x +x^x\ln x$