# Inverse Function Theorem

Could anyone please allude to what this one step proof is for this question:

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Since the function $\,f^{-1}\,$ is derivable with no-where zero derivative, by the theorem about the derivative of the inverse we get

$$f'=\left[\left(f^{-1}\right)^{-1}\right]'=\frac{1}{\left(f^{-1}\right)'}\ldots$$

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Thank you kindly. –  Dick Jan 23 '13 at 15:38

I disagree that this is a one step proof.

You have to prove $f$ is differentiable first to use the chain rule.

@Dicky $f^{-1}$ is differentiable does not directly mean $f$ is differentiable (or even continuous). The only complete proof of this is on the link, any step short is questionable. –  mezhang Jan 24 '13 at 15:35
You observed very well that in the link I sent you, it's a proof of differentiability of $f^{-1}$ knowing $f$ is differentiable and non-zero derivatives, and that your problem is other-wise. But please consider the fact that inverse functions are dual, i.e. if you take $f^{-1}$ as $f$, then $f$ would replace $f^{-1}$ in the proof, and that is what you want. In that sense I sent you the link and expected you to see this. –  mezhang Jan 25 '13 at 1:14