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This is only a half-thought out question right now, and I'll probably answer it myself. But I'm posting it as I came up with it so that, after I work on it, I can check on here and find out how other people approached it.

Okay. So there is a way to find the derivative of a function if you know the derivative of its inverse, like so:

g'(x) = \frac{1}{f'(g(x))} where $f$ is the inverse of $g$.

Now let's say that I know $g$, f' and g' but I don't know $f$. If I have:

f'(g(x))g'(x)=1

Can I solve for f?

If so, how?

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    Even using integrals, you would end up with the equation $f(g(x)) = x+C$, which again provides no information. I just don't see how you can "solve for $f$" without simply solving the "inverse function" equation, $f(g(x)) = x$.2012-03-19

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Basically what you're saying is that you want to solve the differential equation f'(s) = 1/g'(f(s)) for $f(s)$. With a known starting point $f(g(x_0)) = x_0$, the standard numerical methods for solving differential equations will indeed give you numerical approximations for $f(s)$ on an interval where $g$ is continuously differentiable and g'(f(s)) stays away from $0$.

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    Well, one case might be if $g(x)$ is defined as an integral rather than something that is easily computed directly.2012-03-25