Derivative of double integral with respect to upper limits

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.

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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:

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

Hint: Use the following fact:

Let $\phi(\alpha)=\int_{u_1}^{u_2}f(x,\alpha)dx, ~~a\leq\alpha\leq b$, where in the functions $u_1$ and $u_2$ may depend on the parameter $\alpha$. Then $$\frac{d\phi}{d\alpha}=\int_{u_1}^{u_2}f_{\alpha}dx+f(u_2,\alpha)\frac{du_2}{d\alpha}-f(u_1,\alpha)\frac{du_1}{d\alpha}$$

Here we can consider $x$ as $\alpha$ and consider $\int_0^{x}f(y,z)dy$ as $g(x,z)$.

Call your integral $I(x)$. What does $I(x)$ represent? It is the integral of function $f(y,z)$ over a square with one corner at $(0,0)$ and length of side equal to $x$. So that’s $I(x)$. Now, if $x$ goes to $x+\Delta x$, what is $\Delta I$? That is the integral over the area made of two little slivers that wrap around your square from $(y=x, z=0)$, up to $(y=x, z=x)$, and back to $(y=0, z=x)$. So $\Delta I=\Delta x \times (\int_0^xf(x,z)dz + \int_0^xf(y,x)dy)$. Divide $\Delta I$ by $\Delta x$ and you have your derivative.