Possible discrepancy between two forms of the derivative of $|x|^{3/2}$? I've come across what seems to be a discrepancy between two different ways of representing the derivative of the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=|x|^{3/2}$. I started by using the definition of the absolute value function:
$$
f(x) = |x|^{3/2} =
\begin{cases}
x^{3/2}, & x\geq 0 \\
(-x)^{3/2}, & x<0 \\
\end{cases}.
$$
Then using the chain rule, we have
$$
f'(x) = 
\begin{cases}
\frac{3}{2}x^{1/2}, & x>0 \\
-\frac{3}{2}(-x)^{1/2}, & x<0 \\
\end{cases}.
$$
Based on the graph of the function, it seemed reasonable to check if the derivative existed at zero. Thus I made the following computations:
\begin{equation*}
\begin{split}
f_+'(0)&=\lim_{h\to 0^+}\frac{f(h)-f(0)}{h} = \lim_{h\to 0^+}\frac{h^{3/2}}{h}=\lim_{h\to 0^+}h^{1/2}=0 \\
f_-'(0)&=\lim_{h\to 0^-}\frac{f(h)-f(0)}{h} = \lim_{h\to 0^-}\frac{(-h)^{3/2}}{h}=-\lim_{h\to 0^-}\frac{(-h)^{3/2}}{(-h)}=-\lim_{h\to 0^-}(-h)^{1/2}=0.
\end{split}
\end{equation*}
From this, it seems that the derivative exists at $x=0$ and that $f'(0)=0$. However, if we find the derivative using the chain rule and the fact that we can write the derivative of the absolute value function as
$$
\frac{d}{dx}\left[\,|x|\,\right ]=\frac{x}{|x|},
$$
we obtain
$$
f'(x)=\frac{3}{2}|x|^{1/2}\cdot \frac{x}{|x|} = \frac{3x}{2\sqrt{|x|}},
$$
which should be undefined at $x=0$. So the question is, should the derivative be defined at $0$ or should it not? If so, is this just an issue with the notation we use for the derivative of the absolute value function or a sign error or something? Thanks for any input.
 A: We have


*

*for $x>0$
$$f'(x)= \frac{3x}{2\sqrt{|x|}}=\frac{3x}{2\sqrt{x}}=\frac32\sqrt x$$


*

*for $x<0$
$$f'(x)= \frac{3x}{2\sqrt{|x|}}=-\frac{3|x|}{2\sqrt{|x|}}=-\frac32\sqrt {|x|}$$
which agrees with your initial evaluation.
Therefore since $f(x)$ is continuos and 
$$\lim_{x\to 0^+} f'(x)=\lim_{x\to 0^-} f'(x)=0$$
we have that $f'(0)=0$ according to what you have already found directly from the definition.
Refer also to the related: Suppose $f(x)$ exists for all $x\neq0$, and $\lim_{x\rightarrow0}f'(x)$ exists. Show that $f'(0)$ exists.
A: We have $|\frac{f(x)-f(0)}{x-0}|=\sqrt{|x|} \to 0$ as $x \to 0$. Hence $f$ is differentiable at $0$ anf $f'(0)=0$.
A: Your basic error comes from the line where you derive the norm $|\cdot |$ which is not differentiable at $0$. Hence, what you write holds for $x\neq 0$ and you can not conclude like this. The previous calculations are totally okay (and if needed you can shorten then as Fred did in his answer).
A: Gimusi 's link addresses the problem.
This answer also uses the MVT.
$f(x)$ is continuos in $\mathbb{R}$.
$f'(x)$ exists in $\mathbb{R}$ \ {$0$}.,
and 
$\lim_{ x \rightarrow 0^+}f'(x)=$
$\lim_{x \rightarrow 0^-}f'(x)=:L$
Show that $f'(0)$ exists and 
$f'(0) =L$.
MVT:
Let $x >0:$
$\dfrac{f(x)-f(0)}{x} = f'(t)$, $t \in (0,x)$.
$\lim_{x \rightarrow 0^+} \dfrac{f(x)-f(0)}{x}=$
$\lim_{t \rightarrow 0^+}f'(t)=L$.
Similarly for $x <0$, and $x \rightarrow 0^-.$
Hence $f'(0)$ exists and $f'(0)=L$.
A: You can't use the chain rule because
$$|x|'=\frac{x}{|x|}$$ is undefined at $x=0$.
That doesn't mean that the derivative of $|x|^{3/2}$ is undefined, just that you cannot conclude using the chain rule.

The correct way is from the definition of the derivative,
$$\lim_{x\to0}\frac{|x|^{3/2}-0}{x}=\pm\lim_{x\to0}|x|^{1/2}=0.$$
(The $\pm$ choice depends on the sign of $x$, but it doesn't matter.)
