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The Leibniz rule is as follows:

$$\frac{d}{d\alpha} \int_{a(\alpha)}^{b(\alpha)} f(x, \alpha) dx = \frac{db(\alpha)}{d\alpha} f(b(\alpha), \alpha) - \frac{da(\alpha)}{d\alpha} f(a(\alpha), \alpha) + \int^{b(\alpha)}_{a(\alpha)} \frac{\partial}{\partial\alpha} f(x, \alpha) dx$$

What I would like to know is how to apply the above formula for the case of the partial derivative:

$$\frac{\partial}{\partial\alpha} \int_{a(\alpha)}^{b(\beta)} f(x, \alpha) dx.$$

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    What's "it"? Clearly the exact same formula won't hold if you have $b(\beta)$ as the upper limit instead. Which generalization of the result is "it" meant to refer to?2012-11-21
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    @joriki the provided answer seems to suggest that the same formula does hold.2012-11-21
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    The answer shows how to *apply* that formula to your problem. I still don't understand what you might mean by the same formula holding in this case.2012-11-21
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    @joriki What I meant was whether you could *apply* it in the way that `littleO` applied it. Since you can do so, the answer to my question is *yes*. I've edited for disambiguation. Thanks.2012-11-22

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You compute a partial derivative with respect to $\alpha$ by holding $\beta$ fixed, and then just differentiating the resulting function of $\alpha$, which is a function of a single variable. And yes, the Leibniz rule tells you how to differentiate this function of $\alpha$.

For a given $\beta$, the derivative of the function \begin{align*} g(\alpha) &= \int_{a(\alpha)}^{b(\beta)} f(x,\alpha) \, dx \end{align*} is \begin{equation} \frac{dg(\alpha)}{d\alpha} = 0 - \frac{da(\alpha)}{d\alpha} f(a(\alpha),\alpha) + \int_{a(\alpha)}^{b(\beta)} \frac{\partial}{\partial \alpha} f(x,\alpha) \, dx. \end{equation}