## Differentiate under integral sign for iterated integral?

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.

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'as I don't know the answer immediately I thought I'd just ask" ... is this a polite way of saying you've thought about it a bit and can't remember how to do it? (and FWIW it is not like differentiating under the integral sign; moreover, without some conditions on $f$ it's not clear to me that you can say anything precise) – Yemon Choi Apr 1 2010 at 18:25
Also, at the risk of sounding like a grump, did you check the site's FAQ first? mathoverflow.net/faq – Yemon Choi Apr 1 2010 at 18:26
For $f$ continuous, just draw the darn square and ask yourself what the difference is between integrating over $[0,t]^2$ and over $[0,t+{\rm d}t]^2$. – Theo Johnson-Freyd Apr 1 2010 at 21:01

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.

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