Find the derivative of a triple integral function The function $f(x)$ is differentiable.
And I need to find the derivative of the triple integral function:  $$F(t)=\iiint_V \ f(\ xyz \ ) \ dxdydz \ ,
\ \ and \ V=\{(x,y,z)|0\leq x\leq t,\ 0\leq y\leq t,\ 0\leq z\leq t,   \ 0\leq t \}$$ 
I try to let $x=ut,\ y=vt,\ z=wt $
Then the function is
$$ F(t)=\iiint_V \ f(\ uvwt^3\ ) \ t^3dudvdw$$
But I don't know how to do it after that.
 A: Let's start from the Fundamental Theorem of Calculus with a similar problem:
$H(t):=\intop_0^t h(x)dx$
Let's try the limit approach, which requires getting $H(t+\Delta)$:
$H(t+\Delta)=\intop_0^{t+\Delta}h(x)dx=\intop_0^t h(x)dx+\intop_t^{t+\Delta}h(x)dx$
Thus when we take the secant ratio, we get:
$\frac{H(t+\Delta)-H(t)}{\Delta}=\frac{\intop_t^{t+\Delta}h(x)dx}{\Delta}$
The key intuition of the FTC is that when we're very close to $t$ (i.e., $\Delta\approx 0$), $h$ is basically linear, so that it can be approximated with better and better accuracy by $h(x)=h(t)$ on $[t,t+\Delta]$, in which case the integral is just the area of a rectangle with base $\Delta$ and height $h(t)$. That is,
$\frac{H(t+\Delta)-H(t)}{\Delta}\approx\frac{\Delta h(t)}{\Delta}=h(t)$.
Now let's move to the 2D version of the problem with:
$G(t):=\intop_0^t\intop_0^t g(xy)dydx$
$G(t+\Delta)=\intop_0^{t+\Delta}\intop_0^{t+\Delta}g(xy)dydx$
Let's break up the region of integration into four pieces: $[0,t]\times[0,t]$, $[t,t+\Delta]\times[0,t]$,$[0,t]\times[t,t+\Delta]$, and $[t+\Delta,t+\Delta]$. Then $G(t+\Delta)$ is:
$G(t)+\intop_0^t\intop_t^{t+\Delta}g(xy)dxdy+\intop_0^t\intop_t^{t+\Delta}g(xy)dydx+\intop_t^{t+\Delta}\intop_t^{t+\Delta}g(xy)dxdy$
We can approximate the inner integral on the first two terms as above; the last integral is approximated by a square prism with height $g(t^2)$ and square base with side $\Delta$:
$G(t+\Delta)\approx G(t)+\intop_0^t\Delta g(ty)dy+\intop_0^t\Delta g(tx)dx+\Delta^2g(t^2)$
And so the derivative is:
$\frac{G(t+\Delta)-G(t)}{\Delta}\approx \intop_0^t g(ty)dy+\intop_0^t g(tx)dx+\Delta g(t^2)$
The last term disappears as $\Delta\rightarrow 0$, so $G'(t)=\intop_0^t g(ty)dy+\intop_0^t g(tx)dx$
I won't go into the detail for $F$, but you can imagine how it will work out: we break the cube into 8 regions: one is simply $[0,t]^3$, 1 is a cube with side $\Delta^3$, 3 are prisms with one square base $\Delta^2$, and 3 have only one $\Delta$; only these last survive, and the derivative is:
$F'(t)=\intop_0^t\intop_0^t f(txy)dxdy+\intop_0^t\intop_0^t f(txz)dxdz+\intop_0^t\intop_0^t f(tyz)dydz$
We can confirm this is sort of working by setting $f(xyz)=1$; then $F(t)=t^3$, so clearly $F'(t)=3t^2$; it should also be clear that the above formula will generate this as well.
This also works for $f(xyz)=xyz$: $F(t)=(\frac{1}{2}t^2)^3=\frac{1}{8}t^6$, so clearly $F'(t)=\frac{3}{4}t^5$, whereas each term in the above formula will be $t(\frac{1}{2}t^2)^2=\frac{1}{4}t^5$; tripling this gives the desired result.
A: Rescaling the integration variables appropriately, we find:
$$\begin{align}
F{\left(a\right)}
&=\iiint_{V}f{\left(xyz\right)}\,\mathrm{d}V\\
&=\int_{0}^{a}\mathrm{d}x\int_{0}^{a}\mathrm{d}y\int_{0}^{a}\mathrm{d}z\,f{\left(xyz\right)}\\
&=\int_{0}^{a}\mathrm{d}x\int_{0}^{a}\mathrm{d}y\int_{0}^{axy}\mathrm{d}w\,\frac{f{\left(w\right)}}{xy};~~~\small{\left[xyz=w\right]}\\
&=\int_{0}^{a}\mathrm{d}x\int_{0}^{a^2x}\mathrm{d}v\,\frac{1}{xv}\int_{0}^{v}\mathrm{d}w\,f{\left(w\right)};~~~\small{\left[axy=v\right]}\\
&=\int_{0}^{a^3}\frac{\mathrm{d}u}{u}\int_{0}^{u}\frac{\mathrm{d}v}{v}\int_{0}^{v}\mathrm{d}w\,f{\left(w\right)};~~~\small{\left[a^2x=u\right]}\\
&=\int_{0}^{a^3}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\int_{0}^{v}\mathrm{d}w\,\frac{f{\left(w\right)}}{uv}.\\
\end{align}$$
Next, changing the order of integration reduces the triple integral to a single integral:
$$\begin{align}
F{\left(a\right)}
&=\int_{0}^{a^3}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\int_{0}^{v}\mathrm{d}w\,\frac{f{\left(w\right)}}{uv}\\
&=\int_{0}^{a^3}\mathrm{d}u\int_{0}^{u}\mathrm{d}w\int_{w}^{u}\mathrm{d}v\,\frac{f{\left(w\right)}}{uv}\\
&=\int_{0}^{a^3}\mathrm{d}u\int_{0}^{u}\mathrm{d}w\,\frac{\ln{\left(\frac{u}{w}\right)}}{u}\,f{\left(w\right)}\\
&=\int_{0}^{a^3}\mathrm{d}w\int_{w}^{a^3}\mathrm{d}u\,\frac{\ln{\left(\frac{u}{w}\right)}}{u}\,f{\left(w\right)}\\
&=\int_{0}^{a^3}\mathrm{d}w\,\frac{\ln^2{\left(\frac{w}{a^3}\right)}}{2}\,f{\left(w\right)}.\\
\end{align}$$
The derivative of $F{(a)}$ may then be evaluated via the general Leibniz theorem.
