# Manipulating Dirac delta of absolute value function

I am trying to show

$$\delta(2|x|-3)=|x|\delta(x^2-\frac{9}{4}).$$

The composition property states $$\delta(g(x))=\sum\frac{\delta(x-a_i)}{|g'(a_i)|}$$ where $$a_i$$'s are the roots of $$g(x)=0$$.

If I start with the RHS, the roots of $$x^2-9/4$$ are $$\pm 3/2$$ with corresponding derivatives $$\pm 3.$$ This gives $$|x| \delta(x^2-\frac{9}{4})=\frac{|x|}{3}\delta(x-\frac{3}{2})+\frac{|x|}{3}\delta(x+\frac{3}{2}).\tag{*}$$

If I start with the LHS $$2|x|-3$$, has two roots $$\pm 3/2$$ and corresponding derivatives $$\pm 2$$. This implies $$\delta(2|x|-3)=\frac{1}{2}\delta(x+\frac{3}{2})+\frac{1}{2}\delta(x-\frac{3}{2}).\tag{**}$$

Now I am stuck. How do I show that (*) and (**) are equal?

Is it ok to use the composition property for $$2|x|-3$$ even though is not continuously differentiable everywhere?

Simply use the definition: $$\int f(x)\delta(x-x_0)=\int f(x_0)\delta(x-x_0)=f(x_0)$$
Therefore $$|x|\delta(x-x_0)=|x_0|\delta(x-x_0)$$
Define $$f(y) :=y^2-\frac{9}{4}$$. Then $$f(|x|) =|x|^2-\frac{9}{4}$$ and $$f^{\prime}(|x|) =2|x|$$. So
$$|x|\delta(x^2-\frac{9}{4})~=~|x|\delta(f(|x|)) ~=~ \frac{|x|}{|f^{\prime}(|x|)|}\delta(|x|-\frac{3}{2}) ~=~ \frac{1}{2}\delta(|x|-\frac{3}{2}) ~=~ \delta(2|x|-3).$$