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Math Help - Partial derivative using chain rule. Can't find my mistake.

  1. #1
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    Partial derivative using chain rule. Can't find my mistake.

    Evaluate \frac{\partial w}{\partial u} at (u,v)=(1,3) for the function

    w(x,y,z)=xz+yz-z^{2};x=uv,y=uv,z=u

    Ok, so I'm going to to do:

    (\frac{\partial w}{\partial x})(\frac{dx}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du})+(\frac{\partial w}{\partial y})(\frac{dy}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du})

    Right?

    Then I got:

    (z)(v)+(x)(1)+(z)(v)+(y)(1)-(2z)(2u)

    After substituting:

    =>zv+x+zv+y-4uz

    =>uv+uv+uv+uv-4u^{2}|_{(1,3)}

    =>(1)(3)+(1)(3)+(1)(3)+(1)(3)-4(1^{2})=8

    But this isn't one of the answer choices on my exam review. What did I do wrong?
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  2. #2
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    Quote Originally Posted by downthesun01 View Post
    Evaluate \frac{\partial w}{\partial u} at (u,v)=(1,3) for the function

    w(x,y,z)=xz+yz-z^{2};x=uv,y=uv,z=u

    Ok, so I'm going to to do:

    (\frac{\partial w}{\partial x})(\frac{dx}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du})+(\frac{\partial w}{\partial y})(\frac{dy}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du})+(\frac{\partial w}{\partial z})(\frac{dz}{du}) Mr F says: What are the last two terms doing there?

    Right?

    Then I got:

    (z)(v)+(x)(1)+(z)(v)+(y)(1)-(2z)(2u) Mr F says: Please explain clearly where each of these terms have come from. Some of them make no sense at all to me.

    [snip]
    ..
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