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 $y = f(x)$ ?
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 $y = f(x)$ ?
Yes, if f'(x)>0; then (f^{-1})'(y)=1/f'(x). But not if f'(x)=0.
I am afraid that is not true $f= (x-1/2)^2$.
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 $0=f(1/2)$.