TITLE: partial derivative of an integral function
QUESTION [8 upvotes]: Given the function $f(x,y) = \int^x_y g(t) dt $, $g(t)$ continuous for all $t$.
How to evaluate the partial derivatives of $f$ ?
It is correct to do like this (by FTC) ?:
$f_x$ = $g(x)x'$.
$f_y = g(y)y'$.
Thanks!!!

REPLY [4 votes]: "Leibniz's formula" is a generalization of the "Fundamental Theorem of Calculus":  $$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x, t)dt= f(x, \beta(x))- f(x, \alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f(x, t)}{\partial x} dt$$
Here, $f(x,t)$ is a function of t only, the upper bound on the integral is just $x$ and the lower bound just $y$.  So the derivative of the integral with respect to $x$ is $g(x)$ and the derivative with respect to y is $-g(y)$.