I need to prove the idendity of $$\delta(g(x)) = \Sigma \frac{\delta(x-x_i)} {|g'(x_i)|}$$ and we know that $$g(x_i) = 0$$
So I think we can write,
$$g(x) = (x - x_1)(x-x_2)...(x-x_i)$$
so we are actually trying to find,
$$\delta((x - x_1)(x-x_2)...(x-x_i))$$
Now I am not sure what to do. I tried to write $g(x)$ in terms of $g'(x)$ to catch some sort of similarity but it did not work out well.
In another approach, I tried to make a taylor expansion around $x_i$ which gives something like
$$g(x = x_i) = g(x_i) + (x - x_i)g'(x_i) + ...$$
(In here I ignored the higer terms with no reason actually)
Hence around $x_i$, $\delta(g(x)) = \frac{\delta(x-x_i)} {|g'(x_i)|}$ (by using $\delta(g'(x_i) (x-x_i)) = \frac{\delta(x-x_i)} {|g'(x_i)|}$)
since we have $i$ points,
$$\delta(g(x)) = \Sigma \frac{\delta(x-x_i)} {|g'(x_i)|}$$
Is this looks stupid ?