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I am looking to understand how to do differentiation under the integral sign for multiple integrals, as in this article: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign. I have searched online extensively but couldn't find anything that applies.

For example, I want to find the equivalent theorem that can be applied to the following double integral: $$\int_{c_2}^{x_2} \int_{c_1}^{x_1} f(x_1, x_2, t_1, t_2) \,\mathrm{d}t_1 \mathrm{d}t_2 $$ (i.e. I want to differentiate the above by $x_1$ and $x_2$.)

Thank you in advance!

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The essence of the problem is present in problem of differentiating.

$$\int_{1}^{x} f(x, y) \mathrm{d}y $$

with respect to $x$. I'll explain how to do this and leave using it to answer your question as an exercise.

Replace $x$ in the upper limit by $z$. Let $F(x, z)$ denote the integral. The desired derivative will then be $\frac{d}{d x}F(x, z)$ evaluated at $x$.

In order to determine this, we use the chain rule, and then set $z(x) = x$, obtaining

$$\frac{d}{dx}F(x, z) = \frac{\partial F}{\partial x } \frac{d x}{d x } + \frac{\partial F}{\partial z } \frac{d z}{d x } = \int_{1}^{x} \frac{d}{dx} f(x, y) \mathrm{d}y + f(x,y).$$

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Thank you for your help. However, it has been a while since I have done this stuff and I was hoping you could clarify a few things: 1) how did you calculate/reach the evaluation of the chain rule terms (i.e. dF/dx, dF/dx)? 2) To answer the full question, do I simply apply this twice? –  Art Apr 1 '13 at 0:37
    
I'm not really sure I understand how to apply this twice. –  Art Apr 1 '13 at 1:16
    
Chain Rule: math.hmc.edu/calculus/tutorials/multichainrule –  Jonah Sinick Apr 1 '13 at 4:19
    
To answer the full question, let $F(x, y, z, w)$ be the integral that you're considering with the upper limit $x_1$ replaced by $z$ and the upper limit $x_2$ replaced by $w$, and then differentiate with respect to $x$ and $y$ using the multivariable chain rule (which will involve four terms instead of two) and then plug $x_1$ and $x_2$ in at the end. –  Jonah Sinick Apr 1 '13 at 4:22
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