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

$$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}.$$