How to derive below equation?
\begin{equation} \delta(g(x))=\sum_i\frac{\delta(x-a_i)}{\lvert g^\prime(a_i)\rvert}.\qquad (*) \end{equation} I found a solution like equations below: \begin{eqnarray} \int_{-\infty}^{\infty}f(x)\delta(g(x))&=&\sum_i\int_{a_i-\epsilon}^{a_i+\epsilon}f(x)\delta(g(x))\ dx,\qquad g(a_i)=0\\ &=&\sum_i\int_{a_i-\epsilon}^{a_i+\epsilon}f(x)\delta((x-a_i)g^\prime(a_i))\ dx,\qquad (x-a_i)g^\prime(a_i)\ \text{is leading term in its Taylor seires.}\\ &=&\sum_i\int_{a_i-\epsilon}^{a_i+\epsilon}f(x)\frac{1}{\lvert g^\prime(a_i)\rvert}\delta(x-a_i)\ dx,\qquad \delta(ax)=\frac{1}{|a|}\delta(x)\\ \end{eqnarray}
So, $\delta(g(x))=\sum_i\frac{\delta(x-a_i)}{\lvert g^\prime(a_i)\rvert}.$
But comparing equations at fisrt line with third,
$\delta(g(x))=\frac{\delta(x-a_i)}{\lvert g^\prime(a_i)\rvert}.$
What's wrong with above equations? And why use leading term only? eq (*) is approximation?
So my questions are:
When compare with first and third line which are represented below eq (), Where $\sum_i$ in RHS of Eq () come from?
Why use leading term only? eq (*) is approximation? Is it exist that another precise derivation?
Please understand my poor English.