Partial derivative of integral: Leibniz rule?

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:

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

• 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? – joriki Nov 21 '12 at 12:21
• @joriki the provided answer seems to suggest that the same formula does hold. – Jase Nov 21 '12 at 14:07
• 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. – joriki Nov 21 '12 at 17:48
• @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. – Jase Nov 22 '12 at 14:58

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}