# Prove the relation involving derivative of inverse of a function

I want to prove the following result:

$${f^{-1}}'(x)=\frac{1}{f'({f^{-1}}(x))}$$

Is simple application of chain rule a valid proof of it?

i.e. $$f({f^{-1}}(x))=x \implies \frac{df({f^{-1}}(x))}{dx}=1$$ and hence the result. Or is this not a standard proof? Is there additional conditions necessary, expect of course that the function is bijective, or that the inverse exists.

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You also need to apply the chain rule. –  01000100 Feb 27 '13 at 13:03

The question is: Why is $f^{-1}$ differentiable? In the theorem you probably learned in a lecture this isn't assumed. However if you assume that $f^{-1}$ and $f$ are differentiable at and all the expressions make sense $f^{-1}$ should be a function and $f'\neq 0$ (both at least locally).

But nevertheless, with the chainrule the formula is easy to remember.

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For differentiable functions $\mathbb{R} \to \mathbb{R}$ it is easy using geometrical interpretation. Take the graph of $f$, then the graph of $f^{-1}$ is symmetrical with respect to $\{y = x\}$ line. However, $f'(x) = \tan \alpha$ implies that $${f^{-1}}'(y) = \tan\left(\frac{\pi}{2}-\alpha\right) = \cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{f'(x)}.$$
Let $y_0\in \mathbb{R}$ then there's $x_0\in\mathbb{R}$ such that $f(x_0)=y_0$. If $x\rightarrow x_0$ and by continuity of $f$ (because $f$ is differentiable) we have $y=f(x)\rightarrow f(x_0)=y_0$. Now, let's check the differentiability of $f^{-1}$ by the definition: $$(f^{-1})'(y_0)=\lim_{y\rightarrow y_0}\frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=\lim_{y\rightarrow y_0}\frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\\=\lim_{x\rightarrow x_0}\frac{x-x_0}{f(x)-f(x_0)}=\frac{1}{f'(x_0)}=\frac{1}{f'(f^{-1}(y_0))}.$$