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Homework Help: Partial Derivatives

  1. Dec 31, 2007 #1
    [SOLVED] Partial Derivatives

    Whoops, never mind my calculus book explained it.

    1. The problem statement, all variables and given/known data
    F(x,y,z) = 0
    [tex](\frac{\partial x}{\partial y})\right)_{z} (\frac{\partial y}{\partial x})\right)_{z} [/tex] = 0

    [tex](\frac{\partial x}{\partial y})\right)_{z} (\frac{\partial y}{\partial z})\right)_{x} (\frac{\partial z}{\partial x})\right)_{y}[/tex] = -1

    3. The attempt at a solution

    Well I drew out a diagram
    dF/dx dF/dy dF/dz

    dx/dy dx/dz dy/dx dy/dz dz/dx dz/dy

    So I would assume that the reason the second expression is = -1 because of the chain rule, however, I really don't see why it would be -1...
    If it were dx/dy dy/dx dz/dx would it be 1? Just because the middle term is dy/dz the sign will change?
    Last edited: Dec 31, 2007
  2. jcsd
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