Can I apply a Leibniz integration rule to $\frac{d}{dr} \int_{-\infty}^{\infty} \int_{h(r)}^\infty \int_{g(r)}^\infty {rf(x,y,z)\, dz\, dy\, dx}$? Suppose I need to compute the derivative
$$
\frac{d}{dr} \int_{-\infty}^{\infty} \int_{h(r)}^\infty \int_{g(r)}^\infty {rf(x,y,z)\, dz\, dy\, dx}
$$
Can I apply a Leibniz integration rule of some form? How?
 A: Effectively, the below constitutes a derivation of the Leibniz Integral Rule as applicable to the current setting.

Assuming sufficient conditions to allow for interchanging integration and differentiation, and existence of the integrals, you can.
Perhaps the most insightful way is to introduce some auxiliary functions. Define:
$$\begin{align}
G(x,y,b,p) &:= \int_b^\infty pf(x,y,z)\,\mathrm dz \\
H(x,b,c,p) &:= \int_c^\infty G(x,y,b,p)\,\mathrm dy \\
K(b,c,p) &:= \int_{-\infty}^\infty H(x,b,c,p)\,\mathrm dz
\end{align}$$
We want to compute $\dfrac{\mathrm d}{\mathrm dr}K(g(r),h(r),r)$, and start by the general chain rule:
$$\dfrac{\mathrm d}{\mathrm dr}K(b,c,p) = \frac{\partial K(b,c,p)}{\partial b}\frac{\mathrm db}{\mathrm dr}+\frac{\partial K(b,c,p)}{\partial c}\frac{\mathrm dc}{\mathrm dr}+\frac{\partial K(b,c,p)}{\partial p}\frac{\mathrm dp}{\mathrm dr}$$
In this expression, the quantities $\dfrac{\mathrm db}{\mathrm dr},\dfrac{\mathrm dc}{\mathrm dr}$ and $\dfrac{\mathrm dp}{\mathrm dr}$ are easily computed from $b = g(r), c=h(r),p=r$.
For the partial derivatives, we compute as follows:
$$\begin{align}
\frac{\partial K(b,c,p)}{\partial b} &= \int_{-\infty}^\infty \frac{\partial}{\partial b}H(x,b,c,p)\,\mathrm dz\\
&= \int_{-\infty}^\infty \int_c^\infty \frac{\partial}{\partial b}G(x,y,b,p)\,\mathrm dy\,\mathrm dz
\end{align}$$
Now what is $\dfrac{\partial}{\partial b}G(x,y,b,p)$? It is $-pf(x,y,b)$, as we see from the following computation:
$$\begin{align}
\lim_{h\to0} \dfrac{G(x,y,b+h,p)-G(x,y,b,p)}{h} &= \lim_{h\to 0} \frac1h \left(\int_{b+h}^\infty pf(x,y,z)\,\mathrm dz-\int_b^\infty pf(x,y,z)\,\mathrm dz\right)\\
&= \lim_{h\to0} \frac1h \left(-\int_b^{b+h}pf(x,y,z)\,\mathrm dz\right)\\
&= -pf(x,y,b)
\end{align}$$
where in the last step, the Fundamental Theorem of Calculus was used (since the penultimate line comes down to the expression of the derivative of a $z$-primitive of $pf(x,y,z)$ at $z = b$).
The other two partial differentiations can be executed in a similar fashion.
