A question on differentiability of the inverse of strictly monotonically increasing functions

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 $y = f(x)$ ?

• 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
• Not in general: under your assumptions, $f^{-1}$ is differentiable at $y=f(x)$ if and only if $\alpha:=f'(x)\ne0$ (then $({f^{-1}})'(y)=1/\alpha$). – user5317806 May 12 '16 at 14:18

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

• 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 $f= (x-1/2)^2$.

• 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

No. $f(x)=(x−1/2)^3$ is strictly increasing and $f'(x)=3(x-1/2)^2$ and $f'(1/2)=0$, and $g(y):=\sqrt{y}+1/2$ satisfies $g=f^{-1}$ but $g'(0)$ does not exist, as $g'(y)=\frac13 y^{-2/3}$, even though $0=f(1/2)$.