How would you differentiate this? I can't get anywhere Let's say that $F$ is a nice well-behaved function. How would I compute the following derivative?
$\frac{\partial}{\partial t} \left\{ \int_{0}^{t} \int_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) d\xi d\eta \right\}$
I'm guessing I need the fundamental theorem of calculus, but the double integral is REALLY throwing me off - especially that the $t$ is contained in the limits of both integrals. Can someone help me out?
EDIT: Given the problem that this came up in, I have hunch that the above derivative is zero.
 A: First replace the outer integral by $\eta=t$, so the inner integral is $\int_x^x$ which is zero.
Then do the outer integral, but replace the inner integral by $F(x+t-\eta,\eta)+F(x-t+\eta,\eta)$.  
A: I think I figured it out.
$\frac{\partial}{\partial t} \left\{ \frac{1}{2} \int_{0}^{t} \left[ \int_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) d\xi \right] d\eta \right\} \\
= \frac{1}{2} \int_{x - t + t}^{x + t - t} F(\xi,\eta) d\xi + \frac{1}{2} \int_{0}^{t} \frac{\partial}{\partial t} \left\{ \int_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) d\xi \right\} d\eta  \\
= \frac{1}{2} \int_{x}^{x} F(\xi,\eta) d\xi + \frac{1}{2} \int_{0}^{t} \frac{\partial}{\partial t} \left\{ \int_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) d\xi \right\} d\eta  \\
= \frac{1}{2} \int_{0}^{t} \frac{\partial}{\partial t} \left\{ \int_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) d\xi \right\} d\eta \\
= \frac{1}{2} \int_{0}^{t} \left[ F(x - t + \eta,\eta) + F(x + t - \eta, \eta) \right] d\eta \\
= \frac{1}{2} \int_{0}^{t} F(x - t + \eta,\eta) d\eta + \frac{1}{2} \int_{0}^{t} F(x + t - \eta, \eta) d\eta$
