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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.

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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, $f^{-1}(x)=\int_0^x g(t) dt$..

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    @Eelvex: I reverted your edits because (a) the notation you choose is unfortunate: to minimize confusion, as a matter of style, you should not write $(\int g(x) dx)^{-1}(x)$ (b) your use of the indefinite integral is not really an improvement: it is better to just remark that in the last line of niyazi's answer there should be an implicit $+C$.2011-03-13