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Math Help - first partial derivatives

  1. #1
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    first partial derivatives

    Find the first partial derivatives of  f(x,y) = \frac{2x-4y}{2x+4y} at the point (x,y) = (2, 2).

    \frac{\partial f}{\partial x}(2,2) =__
    \frac{\partial f}{\partial y}(2,2)=__
    I need help solving this problem.
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  2. #2
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    Quote Originally Posted by viet View Post
    I need help solving this problem.
    When you take a partial derivatives with respect to a variable, you treat all other variables as constants.

     f(x,y) = \frac{2x - 4y}{2x+4y}

     \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \bigg(\frac{2x - 4y}{2x+4y}\bigg)

    Use the quotient rule:

     = \frac{(2x+4y)\frac{\partial}{\partial x}(2x-4y) - (2x-4y) \frac{\partial}{\partial x}(2x+4y)}{(2x+4y)^2}

     = \frac{(2x+4y)(2) - (2x-4y)(2)}{(2x+4y)^2}


     \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \bigg(\frac{2x - 4y}{2x+4y}\bigg)

    Again use the quotient rule:

     = \frac{(2x+4y)\frac{\partial}{\partial y}(2x-4y) - (2x-4y) \frac{\partial}{\partial y}(2x+4y)}{(2x+4y)^2}

     = \frac{(2x+4y)(-4) - (2x-4y)(4)}{(2x+4y)^2}

    Now plug in the points given in the question.
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