TITLE: Find the derivative of a triple integral function
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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.