The derivative, in the sense of distributions Can you help please to solve this kind of tasks?
enter image description here
u(x) = (3x-1)
where p2(x) is given by 4 options
I tried to find and visited other topics and don't understand what I should do and what is delta(-1) and 1 in answers. I want to know how I can solve this kind of tasks.
I noticed that in which kind of samples I can find only derivate of the function to dx, in which case it will be "3" so I think it is connected with "3" so derivative of the function which equal to 3+p2(x) but the left part where sigmas I don't get it
Thank you very much I think I understood it, but to make sure I want to ask you, does the option b) is correct?
so δ(-1) = -4δ
And δ(1) = 2δ
So I got this -4δ(-1) + 2δ(1) + 3p2(x)
Did I found δ correctly?

If we assume that $p_2(x)$ is defined as
$$p_2(x)=\begin{cases}
1&,-1<x<1\\\\0&,\text{elsewhere}
\end{cases}$$
then for $u(x)=3x-1$, we have in distribution
$$\begin{align}
(u(x)p_2(x))&=u'(x)p_2(x)+u(x)p_2'(x)\\\\
&=3p_2(x)+(3x-1)p_2'(x)
\end{align}$$
where $p_2'$ is the distributional derivative of $p_2$.  Next, we shall analyze the distribution $p_2'$.

Let $\phi \in C_C^\infty$.  Using the definition of the distributional derivative, we assert that
$$\begin{align}
\langle p_2',\phi \rangle&=-\langle p_2,\phi' \rangle\\\\
&=-\int_{-1}^1 \phi'(x)\,dx\\\\
&=\phi(-1)-\phi(1) 
\end{align}$$
Hence, $p2'=\delta_{-1}-\delta_{1}$.

Putting it all together, we have in distribution
$$(up_2)'=(u(-1)\delta_{-1}-u(1)\delta_{1})u+3p_2$$
which can be written colloquially
$$\frac{d}{dx}(u(x)p_2(x))=-2\delta(x-1)-4\delta(x+1)+3p_2$$
