Derivative of the absolute value $|x^3|$, when $x = 0$. Can we find the derivative of $|x^3|$ at $x = 0$?
I was trying to find the derivative of $|x^3|$ within the range of $[-1/2, 1/2]$.
I got the equation for the derivative of $|x^3| = 3x^3 / |x|$. for $x \neq 0$.
Is the equation correct?
And the other thing is I want the derivative of $|x^3|$ at $x = 0$ to obtain the full solution. How can I find it?
Any help would be appreciated.
 A: This is really very simple.
If $x\ge0$, then $f(x)=x^3$ has derivative $3x^2$; so the right derivative at $x=0$ is $0$.
If $x\le0$, then $f(x)=-x^3$ has derivative $-3x^2$; so the left derivative at $x=0$ is $0$.
So the left derivative is equal to the right derivative, and therefore the derivative is their common value, $0$.
A: If $f(x)=|x^3|$, then by the definition of the derivative,
$$
f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{|x^3|}{x} \, .
$$
Since
$$
\lim_{x\to0^+}\frac{|x^3|}{x}=\lim_{x\to0^+}\frac{x^3}{x}=\lim_{x\to 0^+}x^2=0 \, ,
$$
and
$$
\lim_{x\to 0^-}\frac{|x^3|}{x}=\lim_{x\to0^-}\frac{-x^3}{x}=\lim_{x\to0^-}-x^2=0 \, ,
$$
we find that that $f'(0)=0$. Alternatively, we could use the following theorem:

Suppose that $f$ is an even function, and $f'_+(x)$ exists and equals $c$. Then, $f'_-(-x)=-c$.

A: As an alternative, following your first idea, we can use that for $x\neq 0$
$$(|x|)'=\operatorname{sign}(x)$$
then by chain rule and since $x\operatorname{sign}(x)=|x|$
$$f(x)=|x^3|=x^2|x| \implies f'(x)=(|x^3|)'=(x^2|x|)'=2x|x|+x^2\operatorname{sign}(x) =3x|x|=\frac{3x^3}{|x|}$$
and therefore by this theorem
$$\lim_{x\to 0^+}f'(x)=\lim_{x\to 0^-}f'(x) =0 \implies f'(0)=0$$
