## Calculus (3rd Edition)

$$f'(x) =x^{e^x}(e^x\ln x+(e^x/x))$$
Recall that $(e^x)'=e^x$ and $(\ln x)'=\dfrac{1}{x}$. We have $$f(x)=x^{e^{x}}=e^{e^x\ln x}.$$ Now taking the derivative, we get $$f'(x)= e^{e^x\ln x}(e^x\ln x)'=e^{e^x\ln x}(e^x\ln x+(e^x/x))\\ =x^{e^x}(e^x\ln x+(e^x/x))$$