Problem:
Given $g(x)$, solve the equation f'(f^{-1}(x)) = \frac{1}{g(x)} for an invertible and differentiable function $f(x)$.
So far I have tried setting $y = f^{-1}(x) \Leftrightarrow x = f(y)$, obtaining the differential equation f'(y) = \frac{dx}{dy} = \frac{1}{g(x)}
which we then solve to obtain $y$ in terms of $x$, i.e. to obtain the inverse function $f^{-1}(x)$. If this is easily inverted then we can find $f(x)$.
What I am interested in is if anyone knows a better way to solve this, desirably one which allows us to determine $f(x)$ directly.
I'm new to this kind of equation so please correct me if there's a better term for it than "differential-functional" :)
Edit:
I probably should say that in the particular context I am considering, $g(x)$ is the norm of a non-zero vector $\vec{r}(x)$ and is hence always positive.