I have the following function: $ f(x) = \begin{cases} \sqrt{x}, & \text{if $x>0$} \\ \sqrt[3]{|x|}, & \text{if $x<0$ } \end{cases} $.
I have to find $f'(x)$, $f''(x)$ as distributions. With $f'(x)$ it is quite simple: $ f'(x) = \begin{cases} \frac{1}{2\sqrt{x}}, & \text{if $x>0$} \\ -\frac{1}{3\sqrt[3]{|x|}}, & \text{if $x<0$ } \end{cases} $, and there is no problem here as it defines a regular function, since both roots are summable at $0$. I am having trouble in finding $f''(x)$ as $f'(x)$ has some kind of infinite discontinuity, and i don't know how to deal with them. If it was a jump discontinuity, there wouldn't be a problem: we would just get delta function at $0$.
You can just differentiate both parts again to get:
$$ f''(x) = \begin{cases} -\frac{1}{4x\sqrt{x}}, & \text{if $x>0$} \\ -\frac{2}{9\sqrt[3]{x^2}|x|}, & \text{if $x<0$ } \end{cases} $$