Show that the inverse function $ f^{-1} : f(I) \rightarrow I $ is $k$-times continuously differentiable. Let $ k \in \mathbb{N}$. Let $ I \subset \mathbb{R}$ be an open interval. 
Let  $ f : I \rightarrow \mathbb{R}$ be a $k$ times continuously differentiable function  with $f'(x) \not= 0 $ for all $x \in I $.
Show that: 
$1)$ $f$ is injective. 
2) $f(I)$ is an open interval. 
3)  the inverse function $ f^{-1} : f(I) \rightarrow I $ is $k$-times continuously differentiable.
So $1)$ and $2$) weren't a problem. but I need help with $3)$.
I've already shown: $1)$, $2)$, $f^{-1}$ is continuous and $f^{-1}$ is differentiable. 
Many thanks in advance
 A: Let me use $g=f^{-1}$, for ease of notation. By definition, for every $x\in f(I)$,
$$
f(g(x))=x
$$
Differentiating both sides gives
$$
f'(g(x))g'(x)=1
$$
and therefore
$$
g'(x)=\frac{1}{f'(g(x))}
$$
proving that $g'$ is continuous. Suppose $k\ge2$: then we can go further with
$$
g''(x)=-\frac{f''(g(x))g'(x)}{(f'(g(x))^2}
$$
proving that $g''$ is continuous.
Going on this way would be complicated. However, we can use Leibniz's formula: if $F$ and $G$ are $n$ times continuously differentiable functions, then
$$
D^n(FG)=\sum_{i=0}^n\binom{n}{i}D^iF\,D^{n-i}G
$$
and we can use $F(x)=f'(g(x))$, $G(x)=g'(x)$, so for $1\le n\le k-1$,
$$
0=D^n(FG)=\sum_{i=0}^n\binom{n}{i}D^iF\,D^{n-i}G=
g^{(n+1)}+\sum_{i=1}^n\binom{n}{i}D^iF\,D^{n-i}G
$$
and it's a matter of showing that $D^iF$ only depends on derivatives of $f$ up to $i+1$ and of derivatives of $g$ up to $i$, so
$$
g^{(n+1)}=-\sum_{i=1}^n\binom{n}{i}D^iF\,D^{n-i}G
$$
only depends on derivatives of $f$ up to order $n$ and of derivatives of $g$ up to order $n$.
A: Case $k = 1$: $f^{-1}$ is monotonic $\implies$ only can have jump discontinuities. But if the image of an open interval an open interval, no jump discontinuity is possible.
