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Suppose that $f$ is a differentiable function such that $f(g(x)) = x$ and $f^\prime(x) = 1 + [f(x)]^2$. Show that $g^\prime(x) = \dfrac{1}{1+x^2}$.

$$ f(g(x)) = x \implies f = g^{[-1]} $$

I have been told:

$$ f^{[-1]\prime}(f^\prime(c)) = \dfrac{1}{f^\prime(c)} $$

However, I have no intuition of this being true, thus I have a weak understanding of it. I feel as though it may somehow be applicable when considering $f$ is the inverse of $g$ and have been defined as they have. I'm not familiar with the formal name of this truth, so knowing that would be a start on being able to find material from which to study it.

Edit: The problem only defines $f^\prime(x)$. It doesn't define $f^\prime(g(x))$. The formula which was derived is relative to functions, their inverses, and their respective derivatives. It is not necessarily so that $f′(x) = f′(g(x))$. So, how could it be shown that $g'(x) = \dfrac 1 {f'(g(x))} = \dfrac 1 {1+f(g(x))^2} = \dfrac 1 {1+x^2}$ when we don't know what $f^\prime(g(x))$ or $g^\prime(x)$ is?

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2 Answers 2

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Look at the equation $$ f(g(x)) = x$$ and differentiate on both sides, obtaining (via the chain rule) $$f'(g(x)) g'(x) = 1.$$ Thus $$g'(x) = \frac 1 {f'(g(x))} = \frac 1 {1+f(g(x))^2} = \frac 1 {1+x^2}$$

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  • $\begingroup$ Brilliant. So the equation I listed is derived from applying the chain rule? $\endgroup$
    – alxmke
    Mar 17, 2015 at 11:07
  • $\begingroup$ Exactly. It's a neat trick. $\endgroup$
    – Stefan
    Mar 17, 2015 at 11:08
  • $\begingroup$ I'm a little confused, because the problem only defines $f^\prime(x)$ and not $f^\prime(g(x))$. Surely $f^\prime(x) \neq f^\prime(g(x))$, right? $\endgroup$
    – alxmke
    Mar 17, 2015 at 11:25
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Use chain rule ($f=g^{-1}$ or $g=f^{-1}$): $$f(g(x))=x\implies d[f(g(x))]/dx=1\implies f'(g(x)).d[g(x)]/dx=1\implies f'(g(x))g'(x)=1$$

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