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Partial Derivative Chain Rule

  1. Aug 3, 2010 #1
    1. The problem statement, all variables and given/known data
    See figure.

    2. Relevant equations

    3. The attempt at a solution

    Here's what I got,

    [tex]\frac{ \partial z}{\partial x} = \left( \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} \right) + \left( \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} \right) + \left( \frac{\partial z}{\partial x}\right)[/tex]

    Is this correct?

    Attached Files:

  2. jcsd
  3. Aug 3, 2010 #2
    I don't know much about partial derivatives, but it seems weird that what you're solving for (dz/dx), is also your last term on the right side of the equivalency.

    EDIT: Upon further review, looks like I just don't know enough. Sorry for this not-so-helpful post :(
  4. Aug 4, 2010 #3


    User Avatar
    Science Advisor

    Yes, that is correct- although since you are given that u= g(x,y) and v= h(x), it would be better to write
    [tex]\frac{\partial z}{\partial x}= \frac{\partial z}{\partial u}\frac{\partial g}{\partial x}+ \frac{\partial z}{\partial v}\frac{dh}{dx}+ \frac{\partial z}{\partial x}[/tex]
  5. Aug 4, 2010 #4


    Staff: Mentor

    That's exactly what I got, but it bothered me that, since [tex]\frac{\partial z}{\partial x}[/tex] appears on both sides, there doesn't seem to be any way to get an explicit value for this partial.
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