# Derivative of a integral function?

1. Jul 1, 2010

### pellman

How does one work the following?

$$\frac{d}{dx}\int^x_y f(x,u)du$$

I know that (given certain assumptions about the function f)

$$\frac{d}{dx}\int^x_y f(w,u)du=f(w,x)$$

and

$$\frac{d}{dx}\int^c_y f(x,u)du=\int^c_y \frac{df}{dx}(x,u)du$$

but how do we put them together?

2. Jul 1, 2010

### Pere Callahan

$$\frac{d}{dx}\int^x_y f(x,u)du = \int^x_y \frac{d}{dx}f(x,u)du+f(x,x).$$

3. Jul 2, 2010

### pellman

Thanks, Pere.

4. Jul 2, 2010

### HallsofIvy

Staff Emeritus
In general, Lagrange's formula:
$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= f(x,\beta(x))\frac{d\beta}{dx}- f(x, \alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x} dt$$

In this particular problem y is independent of both x and u and can be treated as a constant: dy/dx= 0.