Derivatives of |x|

1. Sep 8, 2014

IniquiTrance

The wikipedia article on $\sgn (x)$ (http://en.wikipedia.org/wiki/Sign_function) states that,

$$\frac{d}{dx}\vert x\vert = \sgn(x)$$

and $\frac{d}{dx}\sgn(x) = 2\delta(x)$. I'm wondering why the following is not true:

\begin{align*} \vert x\vert &= x\sgn(x)\\ \Longrightarrow \frac{d}{dx}\vert x \vert &= \sgn(x) + 2x\delta(x) \end{align*}

by the product rule for derivatives. Is it because this derivative is indeterminate at $x=0$, and $2x\delta(x) \equiv 0$ for all $x \neq 0$?

2. Sep 9, 2014

DavideGenoa

The stated derivative of $\text{sgn}$ is the derivative of $\text{sgn}$ as a distribution, but for distributions that are not identifiable with differentiable functions the chain rule don't apply in general.

Last edited: Sep 9, 2014
3. Sep 10, 2014

jostpuur

It is quite common that formal calculations still give correct results in some sense, even if they are not fully justified.

In this case I don't see anything wrong with the formula

$$\frac{d}{dx}|x| = \textrm{sgn}(x) + 2x\delta(x)$$

Simply subsitute $2x\delta(x)=0$ and you get the previous formula

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

For test function $f$ we have

$$\int\limits_{-\infty}^{\infty} f(x) 2x\delta(x)dx = 0$$

so in this sense $2x\delta(x)=0$ holds "for all $x$", not only for $x\neq 0$.

4. Sep 15, 2014

Xiuh

The formula is true, because $x\cdot \delta$ is actually the zero distribution.