I need to prove the question listed below. It seems like it should be straightforward with a few theorems, but I feel like I am missing something. Was wondering if I could get some second eyes.
Suppose $f(x) = \ln x$ for all $x > 0$ and $g(x) = e^x$ for all real $x$. Given that $f$ and $g$ are inverse functions prove that if $f'(x) = \frac{1}{x}$ for $x \in (0, \infty)$ then $g'(x) = g(x)$ for $x \in (- \infty, \infty)$
Since $f$ and $g$ are inverse functions we have:
$f(g(x)) = x$. Taking the derivative with respect to $x$ and applying the chain rule we have:
$f'(g(x))g'(x) = 1 \rightarrow \ g'(x) = \frac{1}{f'(g(x))}$
Note that as mentioned above $f'(x) = \frac{1}{x}$. Hence, we have:
$g'(x) = \frac{1}{\frac{1}{g(x)}} = g(x)$.
(This might sound hand wavey, but I assure you we have been given a theorem which says this) And we have a theorem which says that if $f$ and $g$ are inverse functions and $f$ is differentiable on $(a, b)$ then $g'(x)$ is differentiable on $(f(a), f(b))$. Thus we have,
$g'(x) = g(x)$ for $x \in (\ln(0), \ln(\infty)) = (-\infty, \infty).$