Possible discrepancy between two forms of the derivative of $|x|^{3/2}$?

Question

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.

Answer 1

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.

Answer 2

We have $|\frac{f(x)-f(0)}{x-0}|=\sqrt{|x|} \to 0$ as $x \to 0$. Hence $f$ is differentiable at $0$ anf $f'(0)=0$.

Answer 3

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).

Answer 4

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 $f'(0)=L$.

Answer 5

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.

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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.)