Differentiation of cos(x)cos(y) or sin(x)sin(y)

[707miranda]

[SOLVED] differentiation of cos(x)cos(y) or sin(x)sin(y)

Homework Statement

Determine if the equation is exact, if exact solve.
[cos(x)cos(y)+2x]dx-[sin(x)sin(y)+2y]=0

Homework Equations

I have forgotten how to take the derivative of cos(x)cos(y) or sin(x)sin(y). Is this a product rule thing, or is it the chain rule?

The Attempt at a Solution

I understand that if
$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
then the equation is exact and you can integrate M w/respect to x and N w/respect to y and combine the similar terms for the solution f(x,y)=solution.

Here $$\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}$$
because the derivative of sin is cos and the derivative of cos is -sin. So while I know the equation isn't exact. I'm still bugged about the whole derivative thing.

Thanks in advance geniuses O o O ><((<))">

 

[cristo]

When you're differentiating a function of two variables, then you treat the one you are not differentiating with respect to as a constant. So, if you want to differentiate sin(x)sin(y) wrt x, then sin(y) is a constant, so this is equal to the derivative of Asin(x), for some constant A. I presume that you know the derivative of this is Acos(x), thus plugging in A gives cos(x)sin(y) as your result.

Use a similar method for the other part: this should tell you whether the equation is exact or not.

 

[Dick]

Follow cristo's advice, the equation IS exact.

 

[Hootenanny]

Hi Miranda and welcome to PF,

I'm guessing that your original equation should be [cos(x)cos(y)+2x]dx-[sin(x)sin(y)+2y]dy=0. You should note that with partial differentiation, assuming that x and y are independent (that is x is not a function of y as visa-versa), you need only differentiate with respect to one variable, whilst holding the other variables constant. So for your example if we take,

\frac{\partial}{\partial x}\left[\cos(x)\cos(y)\right]

Then you would treat y and hence cos(y) as a constant and then differentiate as normal. Hence,

\frac{\partial}{\partial x}\left[\cos(x)\cos(y)\right] = \cos(y)\frac{d}{d x}\cos(x) = -\cos(y)\sin(x)

Do you follow?

Edit: Too damn slow, the roles are reversed cristo !

 

[707miranda]

1,000 thanks I'm not worthy.