TITLE: Derivative of double integral with respect to upper limits
QUESTION [17 upvotes]: How do I perform the following?
$$\frac{d}{dx} \int_0^x \int_0^x f(y,z) \;dy\; dz$$
Help/hints would be appreciated. The Leibniz rule for integration does not seem to be applicable.

REPLY [2 votes]: There is a different way, in my opinion cleaner. Let
$$
F(x,y):=\int_{0}^{x}\int_{0}^{y}f(u,v)\mathop{}\!d v\mathop{}\!d u
$$
Then if we set $F_1(u,y):=\int_0^yf(u,v)\mathop{}\!d v$ then from the fundamental theorem of calculus we have that
$$
\frac{\partial}{\partial x} F(x,y)=\frac{\partial}{\partial x}\int_{0}^x F_1(u,y)\mathop{}\!d u=F_1(x,y)=\int_{0}^{y} f(x,v)\mathop{}\!d v
$$
Similarly, changing the order of integration, we can see that $\frac{\partial}{\partial y}F(x,y)=\int_0^{x}f(u,y)\mathop{}\!d u$. Therefore
$$
\nabla F(x,y)=\left(\int_{0}^{y} f(x,v)\mathop{}\!d v,\int_0^{x}f(u,y)\mathop{}\!d u\right)
$$
Now if you define the function $g:\mathbb{R}\to \mathbb{R}^2,\, t\mapsto (t,t)$ then $(F\circ g)(t)=F(t,t)$, and from the chain rule we finally get that
$$
(F\circ g)'(t)=\nabla F(g(t))\cdot g'(t)=\int_0^{t}f(t,v)\mathop{}\!d v+\int_0^{t}f(u,t)\mathop{}\!d u
$$