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

• 707miranda
In summary, when differentiating a function of two variables, 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.
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 ><((<))">

Last edited:
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.

Last edited:
Follow cristo's advice, the equation IS exact.

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 !

Last edited:
1,000 thanks I'm not worthy.

## 1. What is the formula for differentiating cos(x)cos(y)?

The formula for differentiating cos(x)cos(y) is -sin(x)cos(y)dx - cos(x)sin(y)dy.

## 2. How do you differentiate sin(x)sin(y)?

To differentiate sin(x)sin(y), use the formula cos(x)sin(y)dx + sin(x)cos(y)dy.

## 3. Can the product rule be used to differentiate cos(x)cos(y) or sin(x)sin(y)?

Yes, the product rule can be used to differentiate cos(x)cos(y) or sin(x)sin(y). The product rule states that the derivative of the product of two functions is equal to the first function multiplied by the derivative of the second function plus the second function multiplied by the derivative of the first function.

## 4. Is there a simplified form for the derivative of cos(x)cos(y)?

Yes, the derivative of cos(x)cos(y) can be simplified to -cos(x)sin(y)dx - sin(x)cos(y)dy.

## 5. How can the chain rule be applied to differentiate cos(x)cos(y) or sin(x)sin(y)?

The chain rule can be applied to differentiate cos(x)cos(y) or sin(x)sin(y) by first finding the derivative of the inner function, which in this case is cos(y) or sin(y), and then multiplying it by the derivative of the outer function, which is cos(x) or sin(x), respectively.

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