$f:(a,b) \rightarrow (c,d)$ is a bijection and $f$ is differentibale with $f'(x) \neq 0$ for all $x \in (a,b)$, then $f^{-1}$ is also everywhere differentiable.
Show that if $f$ is twice differentiable then so is $f^{-1}$ and write down the formula for $(f^{-1})''$.
Now my attempt at this question so far,
I do not know how to show that ths $f$ being twice differentiable implies that $f^{-1}$ is also twice differentiable.
But I did attempt the second part of the question
By inverse function theorem we know $$(f^{-1})'(f(x))= \dfrac{1}{f'(x)}$$
now differentiating by chain rule this we get $$f'(x)(f^{-1})''(f(x))=\dfrac{-f''(x)}{(f'(x))^{2}}$$ and rearranging; $$(f^{-1})''(f(x))=\dfrac{-f''(x)}{(f'(x))^{3}}$$
Any hints or help for the first part of th question would be much appreciated.
REMARK; If somebody could advise me as to how to write fractions where the middle line isn't missing, I have tried \frac and \dfrac but both to no success.
Best Answer
Let $f^{-1}=: g$ and denote by ${\rm rec}: z\mapsto {1/z}$ the reciprocal function. It seems you have available the formula $$g'(y)={1\over f'\bigl(g(y)\bigr)}\qquad(c<y<d)\ .\tag{1}$$ This formula can be read as follows: $$g'={\rm rec}\circ f'\circ g\ .\tag{2}$$ On the right side you have a composition of three differentiable functions: ${\rm rec}$ is differentiable wherever its argument $z$ is $\ne0$ (and the derivative is $-{1\over z^2}$); then $f'$ is differentiable by assumption, and $g$ is differentiable according to $(1)$.
We therefore can apply the chain rule to $(2)$ and obtain $$g''=\bigl({\rm rec}'\circ f'\circ g\bigr)\cdot\bigl(f''\circ g\bigr)\cdot g'\ .\tag{3}$$ Writing this out in terms of the variable $y$ and using $(1)$ we obtain $$g''(y)=-{1\over\bigl(f'\bigl(g(y)\bigr)\bigr)^2}\cdot f''\bigl(g(y)\bigr)\cdot{1\over f'\bigl(g(y)\bigr)}=-{f''\bigl(g(y)\bigr)\over\bigl(f'\bigl(g(y)\bigr)\bigr)^3}$$