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I'm stuck in a double integral issue. Actually, my problem it's not with the integral itself, but with a derivative matter. Here is the equation:

\begin{equation} G(z) = \int_{\underline{y}}^{\overline{y}} \int_{0}^{f(z,y)}h(x,y) \ dxdy \end{equation}

I have to compute the first derivative of G(z) with respect to z, but I'm not sure if it's possible to use de Leibniz rule or something else. I'm particulary confuse with the upperboud limit of the second integral, the function f(z,w) (I know the Leibniz rule only when f(.) is one-dimensional).

The h(x,y) function is very non linear and complicated. But my whole objective is to compute g(z):

\begin{equation} g(z) = \frac{d \ G(z)}{dz} \end{equation}

It seems that the solution is relative simple, but I'm not sure :(

Thank you very much.

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Define $$k\left ( t \right )=\underset{D}{\iint} f\left ( x,y,t \right )\mathrm{d}x\mathrm{d}y=\int_{\varphi \left ( t \right )}^{\theta \left ( t \right )}\mathrm{d}x\int_{\tau \left ( x,t \right )}^{\mu \left ( x,t \right )}f\left ( x,y,t \right )\mathrm{d}y$$ let $$p\left ( x,t \right )=\int_{\tau \left ( x,t \right )}^{\mu \left ( x,t \right )}f\left ( x,y,t \right )\mathrm{d}y$$ we get $$\frac{\partial p\left ( x,t \right )}{\partial t}=\frac{\partial \mu \left ( x,t \right )}{\partial t}f\left ( x,\mu \left ( x,t \right ),t \right )-\frac{\partial \tau \left ( x,t \right )}{\partial t}f\left ( x,\tau \left ( x,t \right ),t \right )+\int_{\tau \left ( x,t \right )}^{\mu \left ( x,t \right )} \frac{\partial f\left ( x,y,t \right)}{\partial t}\mathrm{d}y$$ hence $$k'\left ( t \right )=\theta '\left ( t \right )p\left ( \theta \left ( t \right ),t \right )-\varphi '\left ( t \right )p\left ( \varphi \left ( t \right ),t \right )+\int_{\varphi \left ( t \right )}^{\theta \left ( t \right )}\frac{\partial p\left ( x,t \right )}{\partial t}\, \, \mathrm{d}x$$ then the answer will follow.

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