Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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

1 Answer 1

up vote 0 down vote accepted

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 $0$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.