# Computing the derivative: $\frac{\partial}{\partial x} \left\{ \int_0^t \int_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) \,d\xi\, d\eta \right\}$

Let's say that $F$ is a nice well-behaved function. How would I compute the following derivative?

$$\frac{\partial}{\partial x} \left\{ \int\limits_0^t \int\limits_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) \,d\xi \,d\eta \right\}$$

This is what I have so far:

$$= \int_0^t \frac{\partial}{\partial x} \left\{ \int\limits_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) \, d\xi \right\} \, d\eta$$

$$= \int_0^t \left\{ F(x - t + \eta, \eta) + F(x + t - \eta, \eta) \right\} \, d\eta$$

$$= \int_0^t F(x - t + \eta, \eta) \,d\eta + \int_0^t F(x + t - \eta, \eta) \, d\eta$$

Is this correct? Or am I missing a minus sign in there?

Since $$\frac{d}{dx}\int_a^x f(x) \, dx=f(x)$$ and $$\frac{d}{dx}\int_x^a f(x) \, dx=-\frac{d}{dx}\int_a^x f(x)=-f(x)$$ and similarly $$\frac{d}{dx}\int_a^{g(x)} f(x) \, dx=f(x)g'(x)$$
You should have, $$\frac{\partial}{\partial x} \left\{ \int\limits_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) d\xi \right\}= \frac{\partial}{\partial x} \left\{ \int\limits_{x - t + \eta}^{0} F(\xi,\eta) d\xi + \int\limits_{0}^{x + t - \eta} F(\xi,\eta) d\xi \right\}\\ =-F(x - t + \eta,\eta) + F(x + t - \eta,\eta)= F(x + t - \eta,\eta)-F(x - t + \eta,\eta)$$ and therefore $$\frac{\partial}{\partial x} \left\{ \int\limits_{0}^{t} \int\limits_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) d\xi d\eta \right\} =\int_0^t F(x + t - \eta,\eta)-F(x - t + \eta,\eta)d\eta$$