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Given the Leibniz rule:

$$\frac{d}{dy}\int^a_b f(x,y) dx = \int_b^a \frac{\partial f}{\partial y} (x,y) dx$$

How do I prove a more general case using the chain rule and the above:

$$\frac{d}{dy} \int_{g_1(y)}^{g_2(y)} f(x,y) dx =?$$

From the fundamental theorem of calculus we have that: $$\frac{d}{dt} \int_{f_2(t)}^{f_1(t)} g(s) ds = g(f_1(t))f'_1(t) - g(f_2(t))f'_2(t)$$

But when I just apply the fundamental theorem of calculus I get an incorrect answer (because I also need to use Leibniz rule itself...).

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Do the linear change of variables $$x = g_1(y)+(g_2(y)-g_1(y))u.$$

The integral becomes $$\int_{g_1(y)}^{g_2(y)} f(x,y) dx =(g_2(y)-g_1(y)) \int_0^1 f(g_1(y)+(g_2(y)-g_1(y))u,y) du$$ and now you can apply the traditionnal Leibniz Rule.

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