A differential-functional equation: $f'(f^{-1}(x)) = 1/g(x)$

Question

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.

Answer 1

Chen, let $h=f^{-1}.$ Multiply both sides by $h'$ to get $(f'h)h'=h'/g.$ But observe that $f'(f^{-1}(x))(f^{-1})'(x)=(f\circ f^{-1})'(x)=1.$ Hence, we have $(f^{-1})'(x)=g(x).$ So, $f^{-1}(x)=\int_0^x g(t) dt$..