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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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Yup, the answers down below cover it. Just to be sure you're not missing out on anything; the inverse $f^{-1}$ of a real to real function can aquired from the graph of $f$ by just flipping the xy plane along the diagonal. If you haven't already you might find it worthwhile to connect the algebraic picture below with this geometric picture. – Eivind Dahl Apr 25 '11 at 10:09
up vote 3 down vote accepted

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

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By strictly monotonous means does it imply $f'(x) \neq 0 \forall x $ – Rajesh Dachiraju Apr 25 '11 at 10:11
No. For example, $f(x)=x^3$ is strictly increasing, but $f'(0)=0$ nevertheless. – Hans Lundmark Apr 25 '11 at 10:21
It might be added that if you replace "strictly increasing" by just "bijective", then the statement is not true without additional assumptions. (For example, that $f^{-1}$ is continuous at $y$; this is not automatic even if $f$ is bijective and differentiable at $x$, even though it requires a bit of work to come up with a counterexample.) – Hans Lundmark Apr 25 '11 at 10:24
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. – Hans Lundmark Apr 25 '11 at 11:54

I am afraid that is not true $f= (x-1/2)^2$.

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Maybe you meant cubed instead of squared? – Hans Lundmark Apr 25 '11 at 9:58
sqrt is not differentiable in 0 either. cube is correct too. – El Moro Apr 25 '11 at 10:05
The point was that your function is not strictly increasing on $(0,1)$. – Hans Lundmark Apr 25 '11 at 10:25
rightig sorry :) – El Moro Apr 25 '11 at 12:09
so you should edit your answer, rightig? – draks ... Jul 14 '12 at 20:27

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