Why is the derivative of |x| not equal to sgn(x) + 2xδ(x)?

[IniquiTrance]

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

<br /> \frac{d}{dx}\vert x\vert = \sgn(x)<br />

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

<br /> \begin{align*}<br /> \vert x\vert &amp;= x\sgn(x)\\<br /> \Longrightarrow \frac{d}{dx}\vert x \vert &amp;= \sgn(x) + 2x\delta(x) <br /> \end{align*}<br />

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?

 

[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.

 

[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

<br /> \frac{d}{dx}|x| = \textrm{sgn}(x) + 2x\delta(x)<br />

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

<br /> \frac{d}{dx}|x| = \textrm{sgn}(x)<br />

For test function f we have

<br /> \int\limits_{-\infty}^{\infty} f(x) 2x\delta(x)dx = 0<br />

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

 

[Xiuh]

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