Exploring Derivatives of |x| and \sgn(x)

[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 $x \neq 0$?

 

[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

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

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 $x\neq 0$.

 

[Xiuh]

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

 

[blue_raver22]

I would like to clarify that the product rule for derivatives states that for two functions f(x) and g(x), the derivative of their product is given by:

\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)

In this case, we have f(x) = |x| and g(x) = sgn(x), so using the product rule we get:

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

Now, according to the given information, we know that:

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

and

\frac{d}{dx}(sgn(x)) = 2\delta(x)

Substituting these values in the above equation, we get:

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

= 2|x|\delta(x) + (sgn(x))^2

= 2|x|\delta(x) + 1

Now, we can see that the derivative is not equal to just sgn(x) as stated in the original question. This is because the product rule also takes into account the derivative of |x| which is not just sgn(x) but also includes the term 2\delta(x). So, the original statement is not true.

Moreover, as you correctly pointed out, the derivative is indeterminate at x=0, which means that we cannot simply substitute x=0 in the derivative. This is because the derivative of |x| at x=0 is not defined. So, we cannot say that 2x\delta(x) is equal to 0 for all x ≠ 0.

In conclusion, the given statement is not true and the product rule for derivatives should be used correctly to obtain the correct derivative.