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Homework Help: Taking Partial Derivatives

  1. Sep 2, 2008 #1
    1. The problem statement, all variables and given/known data
    If [tex] z = ax^2 + bxy + cy^2 [/tex] and [tex] u = xy [/tex], find [tex] \left(\frac{\partial z}{\partial x}\right)_{y} [/tex] and [tex] \left(\frac{\partial z}{\partial x}\right)_{u} [/tex] .

    2. Relevant equations
    I have Euler's chain rule, the "splitter" and the "inverter" for dealing with partial derivatives.

    3. The attempt at a solution
    I think finding [tex] \left(\frac{\partial z}{\partial x}\right)_{y} [/tex] is easy.
    [tex] \left(\frac{\partial z}{\partial x}\right)_{y} = 2ax + by [/tex]

    However, I do not know how to begin to find [tex] \left(\frac{\partial z}{\partial x}\right)_{u} [/tex] because of the extra function u. One thought is substituting u for xy in the second term on the right side of the original equation ( i wouldn't know how to differentiate it though).

    Any kind of direction would be helpful.

    Thanks in advance.
  2. jcsd
  3. Sep 2, 2008 #2


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    Science Advisor

    Replace y in the function by u/x.
  4. Sep 2, 2008 #3
    All right, let me try your suggestion:

    [tex] z = ax^2 + bx\left(\frac{u}{x}\right) + c\left(\frac{u}{x}\right)^2 [/tex]
    [tex] z = ax^2 + bu + \frac{cu^2}{x^2} [/tex]

    [tex] \left(\frac{\partial z}{\partial x}\right)_{u} = 2ax - \frac{2cu^2}{x^3} [/tex]

    I guess I can resubstitute to get:
    [tex] \left(\frac{\partial z}{\partial x}\right)_{u} = 2ax - \frac{2cy^2}{x} [/tex]

    Is this correct?

    Thanks in advance.
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