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Derivatives of |x|

  1. Sep 8, 2014 #1
    The wikipedia article on [itex]\sgn (x)[/itex] (http://en.wikipedia.org/wiki/Sign_function) states that,

    [tex]
    \frac{d}{dx}\vert x\vert = \sgn(x)
    [/tex]

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

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

    by the product rule for derivatives. Is it because this derivative is indeterminate at [itex]x=0[/itex], and [itex]2x\delta(x) \equiv 0 [/itex] for all [itex]x \neq 0[/itex]?
     
  2. jcsd
  3. Sep 9, 2014 #2
    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
  4. Sep 10, 2014 #3
    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

    [tex]
    \frac{d}{dx}|x| = \textrm{sgn}(x) + 2x\delta(x)
    [/tex]

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

    [tex]
    \frac{d}{dx}|x| = \textrm{sgn}(x)
    [/tex]

    For test function [itex]f[/itex] we have

    [tex]
    \int\limits_{-\infty}^{\infty} f(x) 2x\delta(x)dx = 0
    [/tex]

    so in this sense [itex]2x\delta(x)=0[/itex] holds "for all [itex]x[/itex]", not only for [itex]x\neq 0[/itex].
     
  5. Sep 15, 2014 #4
    The formula is true, because [itex] x\cdot \delta [/itex] is actually the zero distribution.
     
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