1. ## Partial Derivatives Question

I'm having problems finding the partial derivatives in terms of x and y of
$\displaystyle f(x,y)=\int_{y}^{x}\cos(t^{2}) dt$
For some reason I just can't get past the cos(t^2) part and could use some help.

2. Originally Posted by BlueDime
I'm having problems finding the partial derivatives in terms of x and y of
$\displaystyle f(x,y)=\int_{y}^{x}\cos(t^{2}) dx$
For some reason I just can't get past the cos(t^2) part and could use some help.

remember something called the second fundamental theorem of calculus?

it says: $\displaystyle \frac d{dx} \int_c^x f(t)~dt = f(x)$

where $\displaystyle c$ is a constant.

now it's time for a hint: when finding the partial derivative with respect to x, we treat all other variables as constants. a similar holds true for taking the partial derivative with respect to y. i don't suppose i have to tell you this, but i will anyway, $\displaystyle \int_a^b f(x)~dx = - \int_b^a f(x)~dx$

3. Yes I know how to do that part of the problem its just how do you integrate $\displaystyle \int\cos(x^{2}) dx$? I think once I get past there I'll be set.

4. Originally Posted by BlueDime
Yes I know how to do that part of the problem its just how do you integrate $\displaystyle \int\cos(x^{2}) dx$? I think once I get past there I'll be set.
evidently you did not read my post carefully enough. you are not required to find any integral here. look at the theorem again, did it say anything about actually integrating f(t)?

5. Hint: let $\displaystyle f(t) = cos(t^2)$. See? No need to worry about squaring. Or getting an integral at all like the other guy said.

6. Wow I can't believe I let that blow past me. I've got it figured out now. Thanks for the help.