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

  1. Feb 10, 2008 #1
    [SOLVED] differentiation of cos(x)cos(y) or sin(x)sin(y)

    1. The problem statement, all variables and given/known data
    Determine if the equation is exact, if exact solve.

    2. Relevant 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?

    3. 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: Feb 10, 2008
  2. jcsd
  3. Feb 10, 2008 #2


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    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: Feb 10, 2008
  4. Feb 10, 2008 #3


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    Follow cristo's advice, the equation IS exact.
  5. Feb 10, 2008 #4


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    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,

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

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

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

    Do you follow?

    Edit: Too damn slow, the roles are reversed cristo :tongue2:!
    Last edited: Feb 10, 2008
  6. Feb 10, 2008 #5
    1,000 thanks I'm not worthy.
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