Differentiability of $f(|x|)$ at zero This is a self-answered question. I post it here since it wasn't obvious to me. 

Let $f:[0,\infty) \to \mathbb R $ be a function. Then the function $x \to f(|x|)$ is $k$-times differentiable at zero if and only if $f$ is $k$-times differentiable at zero and all the derivatives of $f$ of odd order up to $k$ vanish.

I proved it via (a bit strange) "coupled" induction. Of course, any alternative proofs would be welcomed.
(I feel that there might be a shorter way to phrase this coupling).
To see that the vanishing of the odd order derivatives is necessary, consider the following:
$x \to f(|x|)$ is an even function, so all its odd-order derivatives vanish at zero. However, for non-negative $x$ this map coincides with our original $f(x)$, so its (odd) derivatives should be zero.
 A: We shall prove a stronger claim, using the following (a bit strange) "coupled" induction:
Suppose that $f$ is $k$-times differentiable at zero. Then, 

if all the derivatives of $f$ of odd order up to order $k$ vanish at zero , then
  $\psi: x \mapsto f(|x|)$ is $k$-times differentiable at zero, and if all the derivatives of $f$ of even order up to order $k$  vanish at zero , then $\phi: x \mapsto f(|x|)\text{sgn}(x)$ is  $k$-times differentiable at zero.

Proof:
Suppose that $k=1$. Then
$$ \psi'(0)=\lim_{x \to 0}\frac{\psi(x)-\psi(0)}{x}=\lim_{x \to 0}\frac{f(|x|)-f(0)}{x}=\lim_{x \to 0}\frac{f(|x|)-f(0)}{|x|}\text{sgn}(x)=\pm f'(0)=0,$$
where the $\pm$ refer to the left and right limit. Thus if $f'(0)=0$ then $\psi'(0)$ exists.
Similarly,
$$ \phi'(0)=\lim_{x \to 0}\frac{\phi(x)-\phi(0)}{x}=\lim_{x \to 0}\frac{f(|x|)\text{sgn}(x)}{x}=\lim_{x \to 0}\frac{f(|x|)-f(0)}{|x|}=f'(0),$$
where in the second equality we have used the fact that $f(0)=0$.
Now, suppose that the above claim holds for $k-1$, and let's prove it for $k$.
First, suppose that $f$ satisfies the assumptions for $k$, with the condition on the odd derivatives.
Then
$\psi'(x)= \begin{cases} f'(|x|)\text{sgn}(x) & \text{if $x\neq 0$} \\ f'(0)=0 & \text{if $x=0$}\end{cases} $.
Since $f'$ satisfies the assumptions for $k-1$ (with the condition on even derivatives), it follows that $\psi'(x)$ is $k-1$-times differentiable at zero, so $\psi$ is $k$-times differentiable at zero, as required.
The case where $f$ satisfies the assumptions for $k$, with the condition on the even derivatives is symmetric:
Indeed, in that case $\phi'(x)=f'(|x|)$, and $f'$ satisfies the assumptions for $k-1$ (with the condition on the odd derivatives)- thus it follows that $\phi'(x)$ is $k-1$-times differentiable at zero, so $\phi$ is $k$-times differentiable at zero, as required.
A: $$f(|x|)=\begin{cases}x\ge0\to f(x),\\x\le0\to f(-x)\end{cases}$$
so that
$$f^{(k)}(|x|)=\begin{cases}x\ge0\to f^{(k)}(x),\\x\le0\to f^{(k)}(-x)(-1)^k.\end{cases}$$
Hence all right-derivatives of $f$ must exist and $f^{(k)}(0)=0$ for odd $k$.
A: Note that if the right-hand derivative of $f^{(k)}$ exists at $x=0$ for some integer $k$, then $\psi^{(k)}(0)$ exists, where $\psi:x\mapsto f(|x|).$ In this case, $\psi^{(k-1)}$ is defined in some neighborhood of zero, and  then $\psi^{(k-1)}(x)=f^{(k-1)}(|x|)(\text{sgn} (x))^{k-1}$ whenever $x\neq 0.$ This in turn implies that if $k-1$ is odd, $f^{(k-1)}(0)=0$.
On the other hand, if $\psi^{(k)}$ exists at $x=0,$  then  
$\frac{f^{(k-1)}(|x|)(\text{sgn} (x))^{k-1}-\psi^{(k-1)}(0)}{x}\to \psi^{(k)}(0).$
Now, if $k-1$ is odd then 
$\underset{x\to 0^+}\lim \frac{f^{(k-1)}(|x|)(\text{sgn} (x))^{k-1}-\psi^{(k-1)}(0)}{x}=\underset{x\to 0^+}\lim \frac{f^{(k-1)}(|x|)-\psi^{(k-1)}(0)}{x}$
whereas 
$\underset{x\to 0^-}\lim \frac{f^{(k-1)}(|x|)(\text{sgn} (x))^{k-1}-\psi^{(k-1)}(0)}{x}=\underset{x\to 0^+}\lim \frac{-f^{(k-1)}(|x|)-\psi^{(k-1)}(0)}{x}$
so $f^{(k-1)}(0)=0.$
