# If $f$ is twice differentiable then $f^{-1}$ is twice differentiable

$f:(a,b) \rightarrow (c,d)$ 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.

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The fractions look fine to me! –  String Apr 22 '14 at 14:44

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 $$g''(y)=-{1\over\bigl(f'\bigl(g(y)\bigr)\bigr)^2}\cdot f''\bigl(g(y)\bigr)\cdot{1\over f'\bigl(g(y)\bigr)}=-{f''\bigl(g(y)\bigr)\over\bigl(f'\bigl(g(y)\bigr)\bigr)^3}$$

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## The first order derivatives

For any $y\in(c,d)$ we have a unique $x=f^{-1}(y)\in(a,b)$ and by assumption we know that $f'(x)=\lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}=k$ is well defined and non-zero.

The next part can be phrased more technically via epsilon-delta arguments, but basically we then see that for $\Delta x$ small enough $\Delta y$ is non-zero and we can determine $$(f^{-1})'(y)=\lim_{\Delta y\rightarrow 0}\frac{\Delta x}{\Delta y}=\frac{1}{k}=\frac{1}{f'(x)}$$

## The second order derivatives

Assume now that $f$ is twice differentiable. So \begin{align} (f^{-1})(y+\Delta y)-(f^{-1})(y)&=\frac{1}{f'(x+\Delta x)}-\frac{1}{f'(x)}\\ \color{white}{\frac{0}{0}}\\ &=\frac{f'(x)-f'(x+\Delta x)}{f'(x+\Delta x)\cdot f'(x)} \end{align} The denominator of this tends to $f'(x)^2$ as $\Delta y$ and $\Delta x$ tend to zero and when divided by $\Delta x$ the numerator tends to $-f''(x)$ so we may write $$\frac{f'(x)-f'(x+\Delta x)}{\Delta y}=\frac{f'(x)-f'(x+\Delta x)}{\Delta x}\cdot\frac{\Delta x}{\Delta y}\longrightarrow-f''(x)\cdot\frac{1}{f'(x)}$$ Applying the rules for limits this then eventually leads to the conclusion \begin{align} (f^{-1})''(y)&=\lim_{\Delta y\rightarrow 0}\frac{(f^{-1})(y+\Delta y)-(f^{-1})(y)}{\Delta y}\\ \color{white}{\frac{0}{0}}\\ &=\lim_{\Delta y,\Delta x\rightarrow 0}\frac{\frac{f'(x)-f'(x+\Delta x)}{\Delta y}}{f'(x+\Delta x)\cdot f'(x)}\\ \color{white}{\frac{0}{0}}\\ &=\frac{-f''(x)}{f'(x)^3} \end{align} Just like you wrote.

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