Logarithmic differentiation

Idea. The rule $(x^n)'=nx^{n-1}$ is only for constant exponents. When the exponent varies, rewrite the function as an exponential.

For $f$. $f(x)=e^{x\ln x}$, so
$$f'(x)=e^{x\ln x}\cdot(x\ln x)'=x^x(\ln x+1).$$

For $g$. $g(x)=e^{\sin x\ln x}$, so
$$g'(x)=x^{\sin x}\left(\cos x\,\ln x+\frac{\sin x}x\right).$$

Remark. $f'(x)=0$ exactly when $\ln x=-1$, i.e. $x=\frac1e$: the function $x^x$ has its minimum on $(0,+\infty)$ at $x=\frac1e$, with value $e^{-1/e}\approx0.692$.