About the derivative of the absolute value function

Question

For this question, let $f(x) = |x|$.

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I found this answer saying that the derivative of the absolute value function is the signum function. In symbols,

$$\frac{d}{dx}|x| = \mathrm{sgn}(x).$$

I know that

$$f'(x) = \frac{x}{|x|}$$

using the chain rule. Notice that this is well-defined for $x \neq 0$. However, the definition of the signum function is

$$\mathrm{sgn}\,x = \begin{cases}-1 && \text{for } x< 0 \\ 0 &&\text{for }x = 0 \\ 1 && \text{for } x > 0\end{cases}.$$

This will be my question:

Are $x/|x|$ and $\mathrm{sgn}\,x$ the same derivative of $|x|$?

Answer 1

No they are not. The absolute value function is differentiable at $x$ if and only if $x\ne0$. And, when that happens, the derivative is indeed $\operatorname{sgn}(x)$ or $\frac x{|x|}$. But, since the derivative of the absolute value function is undefined at $0$, but $\operatorname{sgn}$ is defined at that point, those two functions have distinct domains and therefore they cannot possibly be equal.

Answer 2

Yes, in fact you have three common expressions for $\text{sgn}$:

$$\text{sgn}(x) = \frac{x}{|x|} = \frac{|x|}{x}.$$

The two last expressions are of course only valid for $x \neq 0$, but poses no problem when talking about the derivative of $|x|$ since it is not differnetiable at $0$.

EDIT: You can of course not say that the derivative of $|x|$ and $\text{sgn}$ are the same function, since the have different domains. But $\text{sgn}$ provides a very compact way of presenting the derivative of $|x|$ when you make sure that you declare for which $x$ this is valid. What you can write is

$$\frac{d}{dx}|x| = \text{sgn}(x) \quad \text{for } x \neq 0.$$

Answer 3

yes, the function $f(x)=|x|$ is non differentiable at $x=0$ so as far as domain is concerned for $f'(x)$ it is going to be $\mathbb{R} - {{0}}$ and for this domain $x/|x| = sgn(x)$

for better understanding of where functions are non differentiable refer http://www-math.mit.edu/~djk/calculus_beginners/chapter09/section03.html

Answer 4

Summary:

$$|x|'\equiv\frac{|x|}{x}$$ is true.

$$|x|'\equiv\text{sgn}(x)$$ is not.