Derivative of an integral with respect to a shifting region 
Let $f:\mathbb{R}^3 \to \mathbb{R}$ be a smooth function rapidly
  decreasing to zero as $|(x,y,z)| \to \infty$, and let $D(t)$ denote
  $$D(t)=\left\{(x,y,z)\in \mathbb{R}^3 \mid ax+by+cz \leq t\right\},$$
  where $a,b,c\in \mathbb{R}$. Compute $$\frac{d}{dt}
 \int_{D(t)}f(x,y,z)\,dx\,dy\,dz.$$

I'd like to change variables so that the region no longer depends on $t$. Let $$R:=\{(x,y,z)\mid x+y+z\leq 1\}.$$ I note that $$\varphi_t:(x,y,z)\mapsto \left( \frac tax, \frac tb y, \frac tc z \right):R\to D_t$$
Then following the rule $$\int_S \varphi^*(\omega) = \int_{\varphi(S)}\omega$$ I get $$\int_R \frac {t^3}{abc}f\left( \frac ta x, \frac tb y, \frac tc z \right)dx dy dz=\int_{D(t)}f(x,y,z)dxdydz.$$
Now $$\frac{d}{dt}\int_{D(t)}f(x,y,z)\,dx\,dy\,dz = \frac{d}{dt} \int_R \frac {t^3}{abc}f\left( \frac ta x, \frac tb y, \frac tc z \right)dx dy dz$$$$= \int_R \frac{3t^2}{abc}f\left( \frac ta x, \frac tb y, \frac tc z \right) + \frac{t^3}{abc}  \left\langle  \nabla f\left( \frac ta x, \frac tb y, \frac tc z \right), \left( \begin{matrix}t/a \\t/b\\t/c\end{matrix} \right) \right\rangle \tag{1}$$ where we still need to justify switching the integral with the derivative.
Can I simplify that any more, do you think?
To justify the swap of integral and derivative, correct me if I'm wrong, but we would just need the right hand side of (1) uniformly convergent in $t$, for some compact interval of any $t$, right? Can we deduce that from "rapid decay"?
 A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
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 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
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\begin{align}
&\color{#0000ff}{\large%
\totald{}{t}\int_{D\pars{t}}\fermi\pars{x,y,z}\,\dd x\,\dd y\,\dd z}
=
\totald{}{t}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\fermi\pars{x,y,z}\Theta\pars{t - ax - by - cz}\,\dd x\,\dd y\,\dd z
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\fermi\pars{x,y,z}\delta\pars{t - ax - by - cz}\,\dd x\,\dd y\,\dd z
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\fermi\pars{x,y,z}\,{\delta\pars{z - \bracks{t - ax - by}/c} \over \verts{c}}\,\dd x\,\dd y\,\dd z
\\[3mm]&=
\color{#0000ff}{\large{1 \over \verts{c}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\fermi\pars{x,y,{t - ax - by \over c}}\,\dd x\,\dd y}
\end{align}
A: I would like to spell out my comment a little more.  I do not give a complete solution, but hopefully it will give you some ideas, and maybe lead you to the correct solution.  I work in the two dimensional case to make things a little simpler.
Let $I(t)  =\int_{D(t)} f(x,y) dxdy$.  
By the definition of the derivative, $$I'(t) = \lim_{\Delta t \to 0} \frac{1}{\Delta t} (I(t+\Delta t) - I(t)) = \lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{D(t+\Delta t) - D(t)} f(x,y)dxdy$$.
The region $D(t+\Delta t) - D(t)$ is bounded between the two lines $ax+by=t$ and $ax+by = t+\Delta t$.  The perpendicular distance between these two lines is $\frac{\Delta t}{\sqrt{a^2+b^2}}$.
Chop the region into infinitely many squares.  The area of each square is $\frac{\Delta t^2}{a^2+b^2}$.  By the mean value theorem for integrals, the integral we are interested can be rewritten as an infinite sum which looks like 
$$\sum_i \frac{\Delta t^2}{a^2+b^2} f(x_i,y_i)$$
where $(x_i,y_i)$ is an element of each square.
So we are looking at 
$$\lim_{\Delta t \to 0} \frac{1}{\Delta t}\sum_i \frac{\Delta t^2}{a^2+b^2} f(x_i,y_i)$$
Now canceling the $\Delta t$, this is very close to an Riemann sum for an integral of $f$ along the line $ax+by=t$.  It isn't quite, because the $(x_i,y_i)$ are only near the line, not on it.  I think you should be able to use the taylor approximation to $f$ to show that the extra error you get from that fact is negligible in the limit.
My final answer,  (in this 2d case) is 
$$I'(t) = \int_{-\infty}^\infty \dfrac{1}{\sqrt{a^2+b^2}}f(u,\frac{t-ax}{b})du$$ (one factor of \sqrt(a^2+b^2)is absorbed by converting from an integral with respect to arc length to one wrt u).
----------EDIT-----------
Actually I have realized that I am probably reinventing the wheel here.
Stick with the 2 dimensional case for simplicity.
By picking your coordinates appropriately, we may assume that the region is just $y <t$ (we at most get a single constant factor coming from the jacobian of the linear transformation).
This is much easier to inspect.
$$\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{D(t+\Delta t) - D(t)} f(x,y)dydx = \lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{-\infty}^\infty \int_t^{t+\Delta t}f(x,y)dydx$$ 
If you believe we can interchange the limit with the integral here (this is where the real work is) then we have
$$ = \int_{-\infty}^\infty \left[ \lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_t^{t+\Delta t} f(x,y)dy \right] dx$$
But this is the definition of the derivative, and it is being appied to the integral valued function $\int_0^t f(x,y) dy$ (remember $x$ is a constant as far as this integral is concerned).  So the fundamental theorem of calculus then gives
$$ = \int_{-\infty}^{\infty} f(x,t)dx$$.
This agrees with the formula I got above by more heuristic means, and I am sure that writing out the change of coordinates would give you the general case.
Long story short:  my suggestion is to first transform the problem into the case where the "moving plane" is a coordinate axis.  Then (this is the analysis!) see why you can interchange the limits as above.  Apply usual fundamental theorem of calculus, and profit.
It is interesting to consider more complicated $D(t)$.  In this case, since it is just a moving plane, it looks a lot like the normal FTC.  If your integral was WRT an expanding sphere, or a more intricately moving blob, I am not sure I would quite know how to proceed.  It looks like you would maybe get an integral of the interior product of the form you are integrating with the vector field given by the flow of the boundary of region (if you know anything about interior products of differential forms).  If I work out this more general problem I will let you know! 
