Can anyone explain me this result?
Being the derivate of $a\cos(ax)\delta(x)$, where "$a$" is just a constant.
I watched the solution step by step and its result (which is easy to achieve) is: $a(\cos(ax)(\delta(x))'-a\delta(x)\sin(ax))$. But then why does it just become $a(\delta(x))'$?
Well, if we integrate against a test function $f$, we get:
\begin{align*} \int (a\cos(ax)\delta^\prime(x)-a\sin(ax)\delta(x))f(x)dx&=a\int\delta^\prime(x)\cos(ax)f(x)dx-a\int\delta(x)\sin(x)f(x)dx\\ &=a\int\delta^\prime(x)\cos(ax)f(x)dx-\underbrace{a\sin(0)f(0)}_{0}\\ &=a\int\delta^\prime(x)\cos(ax)f(x)dx\\ \text{Integrate by parts}\quad &=-a\int\delta(x)(\cos(ax)f(x))^\prime dx\\ &=-a\int\delta(x)(\cos(ax)f^\prime(x)-a\sin(ax)f(x))dx\\ &=-a\int\delta(x)\cos(ax)f^\prime(x)dx+a^2\int\delta(x)\sin(ax)f(x)dx\\ &=-a\cos(0)f^\prime(0)+\underbrace{a^2\sin(0)f(0)}_{0}\\ &=-af^\prime(0). \end{align*}So the action of $(a\cos(ax)\delta(x))^\prime$ on the test function $f$ is the same as the action of $a\delta^\prime(x)$ (recall that $\delta^\prime$ has the action of returning $-f^\prime(0)$ when acting on the test function $f$).