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I'd like to know if the following statement is true ?

If $f : (0,1) \to \mathbb{R}$ is a strictly monotonically increasing function and $f$ is differentiable at some $x \in (0,1)$ then $f^{-1}(y)$ is differentiable at $y = f(x)$ ?

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    Not in general: under your assumptions, $f^{-1}$ is differentiable at $y=f(x)$ if and only if $\alpha:=f'(x)\ne0$ (then $({f^{-1}})'(y)=1/\alpha$).2016-05-12

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Yes, if f'(x)>0; then (f^{-1})'(y)=1/f'(x). But not if f'(x)=0.

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    Come to think of it, maybe you need to be a little careful here as well... For $f^{-1}$ to be differentiable at $y$, it needs first of all to be defined in a neighbourhood of $y$, so the range of $f$ must contain such an interval, which it doesn't necessarily do (say if $f$ has jump discontinuities accumulating at $x$). But if we add the assumption that $f$ is continuous in a neighbourhood of $x$, then we should be fine.2011-04-25
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I am afraid that is not true $f= (x-1/2)^2$.

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    so you should edit your answer, rightig?2012-07-14
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No. $f(x)=(x−1/2)^3$ is strictly increasing and $f'(x)=3(x-1/2)^2$ and $f'(1/2)=0$, and $g(y):=\sqrt[3]{y}+1/2$ satisfies $g=f^{-1}$ but $g'(0)$ does not exist, as $g'(y)=\frac13 y^{-2/3}$, even though $0=f(1/2)$.