On derivatives of the inverse of a real function

Question

Take two open intervals $I,J\subseteq \mathbb{R}$ and a bijection $f:I\rightarrow J$ with inverse $g:J\rightarrow I$.

For every $k \in \mathbb{N}, k \geq 1$, I know that if $f \in C^k(I,J)$ and for all $t \in I$, $f'(t) \neq 0 $ then $g \in C^k(J,I)$, but I don't know how to prove it. The way I thought I would go about this is:

1. I find the general form of the $k$-th derivative of $g$
2. I realize that $g^{(k)}$ is the composition of continuous functions and is therefore continuous

The trouble is, the general form of $g^{(k)}$ seems to get very complicated, so I was wondering if there is any other way to show this in a simpler way (most likely using induction). My final aim is to prove that $f$ is smooth if and only if $g$ is.

On a related sidenote, I seem to recall that the result is still true if we take $I,J\subseteq \mathbb{R}^n$ open and simply connected (with the jacobian of $f$ being invertible for all $x \in I$). I would just like to know if this is true out of curiousity.

Edit: I forgot an ipotesis on the first derivative being not $0$ on all of $I$, added it.

Answer 1

It is known (e.g. from the inverse function theorem) that $g$ is differentiable and $$ \tag{*} g'(y) = \frac{1}{f'(g(y))} $$ for all $y \in J$, see also Prove the relation involving derivative of inverse of a function for a proof. So $g' \in C^1(J, I)$.

Now assume that $1 \le j \le k-1$ and $g \in C^l(J,I)$. The right-hand side of $(*)$ is a composition of $l$-times continuously differentiable functions and therefore $l$-times continuously differentiable. (This follows e.g. from Faà di Bruno's formula for the higher derivatives of a composite function.) It follows that $g'$ is $l$-times continuously differentiable on $J$, so that $g \in C^{l+1}(J,I)$.

After finitely many steps we get that $g \in C^k(J,I)$.

The same approach generalizes to higher dimensions, where $$ Dg(y) = \bigl[ (Df)(g(y))\bigr]^{-1} $$