The existence of second derivative Let $f(x)=|x|x$ then $f''(0)$ does not exist.
Why?
If $x>0$, $f'(x)=2x$ and if $x<0$, $f'(x)=-2x$.
Then when $x=0$, does $f'(x)$ also not exist?

Write $f$ as:
$$
f(x) = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0\end{cases}
$$
It's easy to see that $f$ is differentiable everywhere. The case $x = 0$ gives $f'(0) = 0$ for the limits from both sides.
We have:
$$
f'(x) = \begin{cases} 2x & \text{if } x \ge 0 \\ -2x & \text{if } x < 0\end{cases}
$$
In other words, $f'(x) = 2|x|$, which is not differentiable at $x = 0$.
