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We know $|x| = \sqrt(x^2)$, determine the second derivative

$\frac{d^2}{dx^2}|x|$,

So the first derivative is sgn(x), but how do I get the second?

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  • $\begingroup$ It doesn't have a derivative using the normal limit definition. It however has a weak derivative, a distribution, normally denoted $\delta$. $\endgroup$ – md2perpe Sep 17 '17 at 12:59
  • $\begingroup$ If $f(x)=|x|$, then $f'(0)$ is undefined. If $x\neq 0$, then $f'(x)=\text{sgn}(x)$ (sign function; signum function). $\endgroup$ – user263326 Sep 17 '17 at 13:19
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HINT:

Consider the graph of $sgn(x)$. What does it look like? What kind of slope does it have?

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  • $\begingroup$ Two horizontal lines from Y = -1 and Y = 1, so the slope is 0 but at X = 0 the slope becomes undefined. True? $\endgroup$ – JohnDoe Sep 17 '17 at 13:49
  • $\begingroup$ Yes, this is not defined in $x=0$. However, you should already know that from the first derivative (where the derivative in $x=0$ is not defined). $\endgroup$ – Niki Di Giano Sep 17 '17 at 13:53
  • $\begingroup$ In fact, the derivative of $|x|$ is not exactly e qual to the signum function, because you already need to restrict the domain to $\mathbb{R} - {0}$. This won't allow you to even consider what happens around $x=0$ to differentiate again, because while it is part of the domain of $sgn(x)$, it is not part of the domain of $\frac{d}{dx} |x|$. $\endgroup$ – Niki Di Giano Sep 17 '17 at 14:17
  • $\begingroup$ Ok I'm lost, so I shouldn't try to get the derivative of sgn(x)? $\endgroup$ – JohnDoe Sep 17 '17 at 14:22
  • $\begingroup$ Actually, you are not left exactly with the $sgn$ function. However, since the only point that is missing in the derivative of $|x|$ is the point at zero, using directly the signum function to differentiate is not that big a problem. You cannot say, though, that the derivative of $|x|$ and the signum function are exactly the same thing, because although they have exactly the same effect in their domains, they do not range over the same domain. $\endgroup$ – Niki Di Giano Sep 17 '17 at 14:31
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You know that

$sgn(x)=\begin{cases} 1 & x>0 \\ 0 & x=0 \\ -1 & x<0 \end{cases}$

I think you can get the derivative from there, derivate each piece of the function. Notice the discontinuity points and consider how this affects the existence of derivative at some points.

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Considering derivative of discontinuity as del(x). Derivative of sgn(x) would be 2*del(x), as there exist a discontinuity at x=0 and a change in step by 2 units (from -1 to +1).

Note : This method is being used in mathematical modeling of signals. Where del(t) is an unit impluse function. And sgn is made up of two step functions.

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