Let me first begin what I mean by saying the intuition behind the " $\delta'(x)$ ". For example the smooth approximations of the delta function looks like the following:
(Left:the smooth approximation of $\delta(x)$
Right:the smooth approximation of $\delta'(x)$)
And by using my intuition I can understand why
$$
\int_{-\infty}^{\infty}f( \bar{x} )\delta(x-\bar{x}) \mathrm{d}\bar{x}=f(x)
$$
because I can say that the delta function fires whenever $x=\bar{x}$ and picks up the value of $f(x)$ at that point and when I integrate over all values of x, I get my function f(x) back. In other words it is like building the function $f(x)$ from thin sticks, which has the same hight as the value of the function. (Although I know that this explanation is nowhere near mathematical, it helps me and others to understand -whatever that means- the concept easier.) When I learned about the derivative of the delta function and its following property I was utterly shocked:
$$
\int_{-\infty}^{\infty}f(\bar{x})\delta'(x-\bar{x}) \mathrm{d}\bar{x}=f'(x)
$$
Because no matter how long I think about the subject I was unable to build a correct intuition about this distribution. My question is this: Can you explain me intuitively why the derivative of the delta function gives arise to a derivative?
PS: I know why this is true mathematically (integrating by parts and so on).
Best Answer
Suppose the spikes in the smooth approximation to $\delta'(x)$ are located at $x=-h$ and $x=h$.
When $\bar{x} \approx x+h$, the smooth approximation to $\delta'(x-\bar{x})$ will be large and positive, so the integral will roughly pick up "something large" times $f(x+h)$. Similary, for $\bar{x} \approx x-h$, the integral will pick up the same large factor times $f(x-h)$, but with the opposite sign. So if that large factor turns out to be of the magnitude $\frac{1}{2h}$, the integral will be roughly $$ \frac{f(x+h)-f(x-h)}{2h} = \frac{\bigl(f(x) + h \, f'(x) + O(h^2)\bigr) - \bigl(f(x) - h \, f'(x) + O(h^2)\bigr)}{2h} , $$ which tends to $f'(x)$ as $h \to 0$.