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Homework Help: Manipulation of partial differential operators.

  1. Dec 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Given that u(x,y) and y(x,z) are both continuous, differentiable functions show that

    ([itex]\frac{\partial u}{\partial z}[/itex])x=([itex]\frac{\partial u}{\partial y}[/itex])x([itex]\frac{\partial y}{\partial z}[/itex])x

    2. Relevant equations

    Only equations given above

    3. The attempt at a solution

    I tried to write out the total differential for du and dy and then holding x constant so dx=0 I rearranged the equations to get

    [itex]\frac{du}{dz}[/itex]=([itex]\frac{\partial u}{\partial y}[/itex])x([itex]\frac{\partial y}{\partial z}[/itex])x

    I then tried to get [itex]\frac{du}{dz}[/itex] in another form but couldn't see a way of getting ([itex]\frac{\partial u}{\partial z}[/itex])x from what I have.

    Any help would be greatly appreciated!
  2. jcsd
  3. Dec 19, 2012 #2


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    Homework Helper

    The chain rule gives
    \left( \frac{\partial u}{\partial z} \right)_x
    = \left( \frac{\partial u}{\partial x} \right)_y
    \left( \frac{\partial x}{\partial z} \right)_x
    + \left( \frac{\partial u}{\partial y} \right)_x
    \left( \frac{\partial y}{\partial z} \right)_x
    So why is it that the first term on the right always vanishes?
  4. Dec 19, 2012 #3
    Sorry mis read that!
    Last edited: Dec 19, 2012
  5. Dec 19, 2012 #4
    I'm sorry I don't know. I thought that the chain rule for partial derivatives was;

    [itex]\frac{du}{dz}[/itex] = ([itex]\frac{\partial u}{\partial x}[/itex])y([itex]\frac{dx}{dz}[/itex]) + ([itex]\frac{\partial u}{\partial y}[/itex])x([itex]\frac{dy}{dz}[/itex])
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