Differentiable function with non-differentiable inverse Is it possible to define a bijective function $f: \mathbb{R} \to \mathbb{R}$ that is differentiable at a point $x_0$ such that $f'(x_0) \ne 0$, but $f^{-1}$ is not differentiable at $f(x_0)$?
I think it is possible and I was trying to split functions for rationals and irrationals but most of those functions fail to be bijective. Any ideas?
 A: Let's construct an example with $x_0=0$. The function $f(x)$  will be equal to $x$ outside of a countable set, namely $C=\{1/n:n\in \mathbb N\} \cup \mathbb N$. 
Define $g:\mathbb N\to\mathbb N$ by $g(n)=n+\lfloor\sqrt{n}-1\rfloor$. This is an injective map which omits infinitely many elements of $\mathbb N$, namely those of the form $k^2+k-2$, $k\ge 2$. Enumerate the omitted integers in the increasing order, $n_1<n_2<\dots$. Also note that $$\lim_{n\to\infty} g(n)/n = 1\tag{1}$$
Define
$$
f(x) = \begin{cases} x, \quad & x\notin C \\ 
1/g(n),\quad & x=1/n, \ \ n\in\mathbb N \\ 
1/n_{k}, \quad & x=2k, \ \ k\in\mathbb N \\
k+1, \quad & x=2k+1, \ \ k\in\mathbb N
\end{cases}
$$
This is a bijection (sends $C$ onto itself and fixes the rest). For example, some values of $f$: 
 x    1   2   3   4    5   6    7    1/2  1/3  1/4  1/5  1/6  1/7  1/8  1/9
f(x)  1  1/4  2  1/10  3  1/18  4    1/2  1/3  1/5  1/6  1/7  1/8  1/9  1/11

By (1) we have $f'(0)=1$. However, $f^{-1}$ is not even continuous at $0$, because $f^{-1}(1/n_{k}) \to\infty $ as $k\to\infty$.
