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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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    Correct me if I'm wrong but I think that's the same result I got as posted in the question, just obtained using a different method.2011-03-12
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    @Josh: I upvoted your comment, but this answer does show that if $g(t)\gt 0$ there is an inverse, which your question does not make explicit.2011-03-12
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    Josh:No, they are not the same. However, one should be more careful, for instance, i am multiplying both sides by $f^{-1}(x)$ which means that I am assuming it is not zero, also as noted by Josh, integral of g increasing..etc2011-03-12
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    @Ross: Sorry I can't see how $g(x)>0$ implies the existence of an inverse. Could you explain? But it was helpful you mentioned this because in the particular context I am looking at this problem, $g(x)$ _is_ always positive - in fact it's a norm of a vector. So if I understood you correctly this would mean that the equation always has a solution $f(x)$?2011-03-13
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    @niyazi: I'm going to be a bit of a pain and ask you elaborate on the "integral of g increasing.." part. Thanks :)2011-03-13
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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