First and second derivative of $|x|^3$ I need to prove that $|x|^3$ is twice differentiable, by showing that the first and second derivatives exist using the definition. I've tried several ways, this is what I've got: $$\lim_{h\to 0} \frac{|x+h|^3 - |x|^3}{h} = \lim_{h\to 0}\frac{\sqrt{(x+h)^2}^3 - \sqrt{x^2}^3}{h} = \lim_{h\to 0} \frac{\sqrt{(x+h)^6} - \sqrt{x^6}}{h}$$ then I rationalized the numerator: $$= \lim_{h\to 0} \frac{(x+h)^6 - x^6}{h\left(\sqrt{(x+h)^6} + \sqrt{x^6}\right)}$$ and I'm stuck on what to do next, I'm skeptic that this is the right way, but I was not able to reach any answer using other ways either. Any help would be great. Thanks.

At $x\
e0$, you can take $h$ so small that $x$ and $x+h$ have the same sign. Then the function can be processed as two polynomial pieces which derive seamlessly:

$$|x^3|=\pm x^3$$ gives the derivatives $$\pm3x^2=3x|x|$$ and $$\pm6x=6|x|$$ (as the $\pm$ signs match that of $x$).

At $x=0$, the first derivative is

$$\lim_{h\to0}\frac{|h^3|}{h}=\lim_{h\to0}\pm h^2=0$$

(the sign doesn't matter as convergence is to zero). Note that the first derivative is continous as $3x|x|$ evaluates to $0$.

The second derivative is given by

$$\lim_{h\to0}\frac{3h|h|}{h}=\lim_{h\to0}3|h|=0.$$

* * *

Note that for the third derivative we get

$$\pm6$$ where the sign is that of $x$, and for $x=0$,

$$\lim_{h\to0}\frac{6|h|}{h}$$ is not defined.