Two questions concerning dirac delta function and Hamiltonian

Question

I'm trying to compute to quantities with Hamiltonian and Dirac delta function but I don't how to do it properly. I'm stuck calculating the following quantity

$$ \frac{d}{dE} \left[ \theta(E-H(x,p;V)) \right]. \tag{1}$$ I know that the derivative of the Heaviside step function $\theta(x)$ is the dirac delta function $\delta(x)$ $$ \frac{d\theta(x)}{dx}= \delta(x)$$

So my guess would be the following:

$$ \frac{d\theta(E-H)}{dE}= \delta(E-H).$$

However if I use the chain rule I get: $$ \frac{d\theta(E-H)}{d(E-H)}\frac{d(E-H)}{dE}= \delta(E-H) \left(1-\frac{dH}{dE}\right).$$

The text I'm studying about areas in phase planes gives the following but leaves out a lot of intermediate steps:

$$ \frac{d}{dE} \left[ \theta(E-H(x,p;V)) \right]=\delta (p^2/2m+\phi-E).$$

Where $H=p^2/2m+\phi(x;V).$

The next step makes me equally confused, it gives the following equation:

$$ \int dp \delta (p^2/2m+\phi-E)=2m/p(x) \ \ \ \ \ \ \ (2)$$

Where the text tells it uses the formula $$\delta(f(p)) = \sum_i \delta(p-p_i)/|f'(p_i)|.$$

Questions How to understand the following equations:

$$ \frac{d}{dE} \left[ \theta(E-H(x,p;V)) \right] =\delta (p^2/2m+\phi-E)) \ \ \ \ \ \ \ \ \ \ (1)$$

$$ \int dp \delta (p^2/2m+\phi-E)=2m/p(x) \ \ \ \ \ \ \ (2)$$

Answer 1

The first is simple I think: $$\frac{d}{dE}\theta[E-H(x,p;V)]=\delta[E-H(x,p;V)]=\delta[H(x,p;V)-E]= \delta\Big[\frac{p^2}{2m}+\phi(x;V)-E\Big]$$ where in the third step I used the symmetric property of the $\delta$ function, i.e. $\delta(-x)=\delta(x)$.

As far as the second question is concerned, one can proceed in the following way $$\int dp\delta(p^2/2m+\phi-E)=\int dp \Big[\frac{1}{|p_+(x)/m|}\delta(p_+^2/2m+\phi-E) +\frac{1}{|p_-(x)/m|}\delta(p_-^2/2m+\phi-E)\Big]$$ where $p_{\pm}=\pm\sqrt{2m(E-\phi)}$ are the two solutions to the equation $f(p)=p^2/2m+\phi-E=0$. The $x$ dependence of $p$ indicates that $p$ is a function of $x$ through $\phi$. Performing the integrals and identifying that the magnitude of $p_+(x)$ equals the magnitude of $p_-(x)$, yields $$\int dp\delta(p^2/2m+\phi-E)=\Big[\frac{1}{|p_+(x)/m|} +\frac{1}{|p_-(x)/m|}\Big]=\frac{2m}{p(x)}$$ where $p(x)=|p_+(x)|=|p_-(x)|=\sqrt{2m(E-\phi)}$.

I hope this helps! If anything is unclear, please comment!