How to prove this Dirac Delta Function property? How to prove the equation below, using Dirac Delta function properties?
$$
\delta(x^2-m^2)=\frac{1}{2|w|}(\delta(x-w)+\delta(x+w))
$$
where 
$$
w^2=|x|^2+m^2
$$
I tried to show it using 
$$
\delta(f(x))=\sum\frac{\delta(x-x_i)}{|f'(x_i)|}
$$
but could not success.
 A: The starting point is the definition:
$$
\delta(f(x))=\sum_i\dfrac{\delta(x-x_i)}{|f'(x_i)|}
$$
where $x_i$ are all the roots of $f(x)$
( the formula in OP is not correct)
For $f(x)=x^2-y^2$ the roots are $x_1=-y$ and $x_2=y$ and $f'(x)=2x$, so we have:
$$
\delta(x^2-y^2)=\dfrac{\delta(x-y)}{2|+y|}+\dfrac{\delta(x+y)}{2|-y|}=\dfrac{\delta(x-y)+\delta(x+y)}{2|y|}
$$
A: I think I got it. As Emilio mentioned I will start with the definition.
$$
\delta(f(x))=\sum_{i}\frac{\delta(x-x_i)}{|f'(x_i)|}
$$
$$
\delta(x^2-m^2)=\frac{\delta(x-m)+\delta(x+m)}{2|m|}
$$ 
with the transformation $w^2=x^2+m^2$
$$
\delta(x^2-m^2)=\frac{1}{2|m|}(\delta(x-\sqrt {w^2-x^2})+\delta(x+\sqrt{w^2-x^2}))
$$
Let me derive the first and second term, using similar definition 
$$
\delta(f(x))=\frac{\delta(x-x_0)}{|f'(x_0)|}
$$ 
$$
\delta(x-\sqrt {w^2-x^2})+\delta(x+\sqrt{w^2-x^2})= \frac{\delta(x-w)}{\frac{2|w|}{2|m|}}+\frac{\delta(x+w)}{\frac{2|w|}{2|m|}}
$$
and eventually come to the end
$$
\delta(x^2-m^2)=\frac{1}{2|w|}(\delta(x-w)+\delta(x+m))
$$
