[Math] Derivative of double integral with respect to upper limits

Question

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.

Answer 1

The Leibniz integral rule does work.

$$\begin{align}\frac{d}{dx} \int_{0}^{x} dy \left[\int_{0}^{x} f(y,z) dz \right] &=\int_{0}^{x} f(x,z) dz + \int_{0}^{x} dy \frac{\partial}{\partial x}\left[\int_{0}^{x} f(y,z) dz \right]\\ &= \int_{0}^{x} f(x,z) dz + \int_{0}^{x} dy \left[f(y,x) + \int_{0}^{x} \frac{\partial f(y,z)}{\partial x} dz\right]\\ &= \int_{0}^{x} f(x,z) dz + \int_{0}^{x} f(y,x) dy \end{align}$$ You only need to remember when the integration limit depends on $x$, $\frac{d}{dx}$ on the integral will pick up extra terms for the integration limits. In general:

$$\frac{d}{dx} \int_{a(x)}^{b(x)} g(x,y) dy = g(x,b(x)) b'(x) - g(x,a(x)) a'(x) + \int_{a(x)}^{b(x)} \frac{\partial g(x,y)}{\partial x} dy$$