Finding the weak derivative of order $3$ of $f(x)=\operatorname{sgn} \sin(x)$ where $\operatorname{sgn}$ is the sign function

Question

Let $$f(x)=\operatorname{sgn} \sin(x)$$ where $\operatorname{sgn}$ is sign function. I need to find the weak derivative of order 3 for $f(x)$?

Do you know this approximation?: $\operatorname{sgn}(x) \approx \dfrac{x}{\sqrt{x^2+\epsilon}}$, where $\epsilon \to 0$ — Inceptio, Apr 30 '13 at 14:52
@ Lost Yes, $f(x)=sgn(sin(x))$ or $f(x)= sin(x)$ for $x>0$,$f(x)=-sin(x)$ for $x<0$ and $f(x)=0$ for $x=0$. — hermman, Apr 30 '13 at 15:03
@Inceptio I don't know this approximation! is this useful for my question? — hermman, Apr 30 '13 at 15:06
I'm hinting so. In fact it is the derivative of $\sqrt{x^2+\epsilon}$. — Inceptio, Apr 30 '13 at 15:07

score 1 · Answer 1 · answered Apr 30 '13 at 19:09

I'm not convinced that approximation of $\mathrm{sgn}\,x$ by $x/\sqrt{x^2+\epsilon}$ makes life easier. Some points to make:

The function $f$ is locally constant on $\mathbb R\setminus (\pi \mathbb Z)$. Therefore, its derivative on this open set is zero.
The distributional derivative is local: its restriction to an open interval such as $(\pi n-\pi/2, \pi n+\pi/2)$ is determined only by the values of $f$ on that interval.
You should know the distributional derivative of the Heaviside (step) function.
But if you don't, it's the Dirac delta, which acts on test functions by evaluating them at a point.
When you take derivatives of a distribution $f$, you pass them to the test functions $\varphi$ via $\langle f',\varphi\rangle =\langle f,-\varphi'\rangle$. This is why, for example, the derivative of the Dirac delta acts on test functions by evaluation of $-\varphi'$ at a point. And so on.

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