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Problem finding a partial derivative

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data

    I am working on a homework problem involving partial derivatives. I've been checking my answers against what Wolfram Alpha spits out just for extra assurance. For the following problem

    Find all the second partial derivatives: v = [itex]\frac{xy}{(x-y)}[/itex].

    When I get to the point where I am deriving v[itex]_{xy}[/itex], Wolfram Alpha gives me

    [itex]\frac{2xy}{(x-y)^3}[/itex]

    but I get something different. If someone can tell me where I am going wrong here, I would be most grateful: I've stared at this for some time now. I suspect it is something very trivial. I am using the quotient rule.

    2. The attempt at a solution

    [itex]\frac{\partial}{\partial y}[/itex] [itex]\frac{y^2}{(x-y)^2}[/itex]

    [itex]\frac{(x-y)^2 * 2y - y^2 * (2 (x-y))}{(x-y)^4}[/itex]

    [itex]\frac{(x-y)((x-y)*2y - 2y^2)}{(x-y)^4}[/itex]

    [itex]\frac{2xy-4y^2}{(x-y)^3}[/itex]

    Thanks,

    Glenn
     
  2. jcsd
  3. Oct 9, 2011 #2
    You forgot the minus.
    [itex]\frac{\partial}{\partial y}[/itex] [itex]\frac{-y^2}{(x-y)^2}[/itex]

    And in your quotient rule, you forgot the derivative on the "inner" function, which is -1.

    [itex] \frac{\partial(x-y)^{2}}{\partial y}=2(x-y)(-1)[/itex]
     
  4. Oct 9, 2011 #3
    Ahh, I was even further off than I thought. Thanks so much. It makes sense now.
     
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