This is a bit of a trivial question, but as I don't know the answer immediately I thought I'd just ask.
Given the integral $\int_{0}^{t} \int_{0}^{t} f(x,x') dx dx'$, what is $\frac{\partial}{\partial t} \int_{0}^{t} \int_{0}^{t} f(x,x') dx dx'$? It looks a bit like differentiating under the integral sign, but I'm not sure how to handle it.
As usual when differentiating something with respect to a variable that appears twice. The chain rule for partial derivatives.
For example, consider function $z = f(u,v)$. Suppose we want $(d/dt)f(t,t)$. Let $u=v=t$ and use $dz/dt = (\partial z/\partial u)(du/dt) + (\partial z/\partial v)(dv/dt)$.
Thus… $$ \frac{d}{dt}\int_0^t\int_0^t f(x,y)\,dx\,dy = \int_0^t f(t,y)\,dy + \int_0^t f(x,t)\,dx $$
By the way, why did you write $\partial/\partial t$ to differentiate a function of the single variable $t$? It's not wrong, just confusing to students.
$f$continuous, just draw the darn square and ask yourself what the difference is between integrating over$[0,t]^2$and over$[0,t+{\rm d}t]^2$. $\endgroup$