TITLE: Partial derivative of integral: Leibniz rule?
QUESTION [7 upvotes]: 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:
$$\frac{\partial}{\partial\alpha} \int_{a(\alpha)}^{b(\beta)} f(x, \alpha) dx.$$

REPLY [3 votes]: 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}