Let $f(x)=|x|^3$. Calculate $f'(x), f''(x)=(f')'(x)$ and show that $f''(x)$ is not differentiable. So, I've got this problem, but I'm not able to finish it.
What I've got until now, is that
$$f (x)=f (x_0)+(x+x_0)f'(x_0)$$
And if $ f $ is differentiable in $x_0$ with derivate $ f'(x) $
$$\Delta x:=\frac{f (x)-f (x_0)}{x-x_0}=\frac {\Delta f (x)}{\Delta x}$$
And doing $x\to x_0\leadsto f'(x)$
I tried and tried and, wondering which point I lost, I got here (not so far)
$$|x|^3=|x_0|^3=(x-x_0)f'(x)$$
$$ f'(x)=\frac {|x|^3-|x_0|^3}{x-x_0}$$
$$\lim_{x\to x_0}f'(x)$$
Probally, we'll got the answer here. But I just obtain $\frac{0} {0} $. I think I forgot all my calculus class and I can't make it.
Please, a little help.
 A: Notice that $f$ may be explicitly written as
$$f(x) = \begin{cases}
x^3, & x>0 \\
0, & x=0 \\
-x^3, & x<0 .
\end{cases} $$
It follows that the only problems are in $0$, because $x^3$ and $-x^3$ are obviously differentiable.
For the first derivative we have that
$$0 \le |f'(0)| = \left| \lim _{x \to 0} \frac {f(x) - f(0)} {x - 0} \right| = \lim _{x \to 0} \left| \frac {f(x) - f(0)} {x - 0} \right| = \lim _{x \to 0} \frac {|f(x)|} {|x|} = \lim _{x \to 0} \frac {|x|^3} {|x|} = \lim _{x \to 0} |x|^2 = 0 ,$$
where the modulus has jumped inside the limit because the modulus function $| \cdot | : \Bbb R \to \Bbb R$ is continuous. We have shown that $0 \le |f'(0)| \le 0$, whence it follows that $f'(0) = 0$, so we may write
$$f'(x) = \begin{cases}
3x^2, & x>0 \\
0, & x=0 \\
-3x^2, & x<0 .
\end{cases} $$
With exactly the same type of argument as above, show for yourself that
$$f''(x) = \begin{cases}
6x, & x>0 \\
0, & x=0 \\
-6x, & x<0 .
\end{cases}$$
Clearly, $f''$ is differentiable on $(-\infty, 0) \cup (0, \infty)$. Let's see what happens in $0$. Choosing to approach $0$ from the left, we must use $f''(x) = -6x$, so
$$\lim _{x \to 0, \ x<0} \frac {f''(x) - f''(0)} {x-0} = \lim _{x \to 0, \ x<0} \frac {-6x} {x} = -6 ,$$
while choosing to approach $0$ from the right forces us to use $f''(x) = 6x$, leading to
$$\lim _{x \to 0, \ x>0} \frac {f''(x) - f''(0)} {x-0} = \lim _{x \to 0, \ x>0} \frac {6x} {x} = 6 .$$
Given that the two lateral limits differ, we conclude that $\dfrac {f''(x) - f''(0)} {x-0}$ has no limit in $0$, which means that $f''$ is not differentiable in $0$.
