# Can I find the inverse of a function using this formula?

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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If you know $g$, then $f$ is the inverse of $g$; you don't need $f'$ nor $g'$ to "find" $f$; and $f'$ and $g'$ may provide absolute no new information; e.g., if $g(x) = x-a$, (so $f(x) = x+a$), then $g'(c)=f'(c) = 1$ for all $c$, so $f'(g(x))g'(x)=1$ just says $1\cdot 1 = 1$; this doesn't tell you what $f$ and $g$ are. – Arturo Magidin Mar 19 '12 at 18:24
I know that f is the inverse of g..........-_- My question was: can I use what is known above to find f? Assume that I don't just know the inverse of g already, obviously. That would pretty much render my question moot wouldn't it. – Korgan Rivera Mar 19 '12 at 18:32
And if you see the second part of the comment, you'll see that there are infinitely many functions $f$ and $g$ that give you the exact same equation $f'(g(x))g'(x)=1$, namely, $f(x)=x+a$, $g(x)=x-a$ with $a$ an arbitrary constant (for which the equation reduces to $1\times 1 = 1$). So this equation need not provide you with any information whatsoever. – Arturo Magidin Mar 19 '12 at 18:34
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, $f(g(x)) = x$. – Arturo Magidin Mar 19 '12 at 18:40

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 $0$.
Numerical methods for solving the non-differential equation f(g(x)) = x should usually work at least as well, and be a lot simpler; such methods are indeed commonly used in practice. There may be cases where it is more effective to solve $f^\prime(s) = 1/g^\prime(f(s))$ instead, but if so I wonder what they are. – MJD Mar 19 '12 at 19:14
Well, one case might be if $g(x)$ is defined as an integral rather than something that is easily computed directly. – Robert Israel Mar 25 '12 at 5:30