# Finding Partial Derivative of an Integral

1. Mar 5, 2008

### physman88

Hey everyone. Im new here and i have a problem with some partials. We're studying partial derivatives in calculus III. I understand and all, but we haven't covered how to take a partial derivative of an integral. This problem showed up in my practice problems before our exam tomorrow.

The problem is as follows:

$$\frac{\partial}{\partial}x$$$$\int$$cos(t$$^{3}$$)dt

If you cant follow that.. then it says we need the first partials (x and y) of the integral of cos(t^3)dt.. (lower limit=y, and upper limit=x)

Any insight on how to start this problem?? Thanks for any help!!

-Kev

2. Mar 5, 2008

### HallsofIvy

Staff Emeritus
I couldn't follow it because you didn't put in the limits of integration, and that is crucial!

You should know from single variable calculus, the "Fundamental Theorem of Calculus":
$$\frac{d}{dt}\int_a^x f(t)dt= f(x)$$
where a is any constant.
From that it should be easy to find the partial derivative with respect to x.

To find the derivative with respect to y, remember that
$$\int_y^a f(t)dt= -\int_a^y f(t)dt$$

3. Mar 5, 2008

### physman88

I understand that you couldn't follow it, but may I ask how to get the limits on the integral? Sorry for the mix-up.

4. Dec 9, 2008

### fcpeace17

HallsofIvy...your d/dt should be a d/dx for the fund. theor.... the way you have it written it would be zero

5. Dec 9, 2008

### tiny-tim

Hey Kev!

Hint: start by thinking ∫f = g …

then ∂/∂x of (∫f between y and x)

= ∂/dx ([g] between y and x).