Is this sufficient to prove that $xe^x$ is invertible and its inverse is differentiable from $(0,\infty)$?
Solution 1:
Let $x\in(0,\infty)$. Since $f'(x)>0$, by inverse function theorem, $f$ is a local diffeomorphism. To find the derivative, using chain rule, differentiate
$$x=f(f^{-1}(x))$$
giving
$$1=f'(f^{-1}(x))(f^{-1})'(x)$$
so
$$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}.$$