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Math Help - Finding the partial derivatives using the chain rule:

  1. #1
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    Finding the partial derivatives using the chain rule:

    Find the indicated derivative by using the chain rule:
    Find du/dt & du / dz not ds
    Last edited by ^_^Engineer_Adam^_^; March 7th 2007 at 02:26 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the indicated derivative by using the chain rule:
    i think you mean you want du/dt and du/dz. All the d's here are partial derivatives, so you will write them as you see in the problem.

    by the chain rule:

    du/dt = du/dx*dx/dt + du/dy*dy/dt

    => du/dt = (ycos(xy))(2z*e^t) + (xcos(xy))(2t*e^-z)
    => du/dt = [t^2*e^-z(cos(2z*e^t*t^2*e^-z)][2z*e^t] + [2z*e^t(cos(2z*e^t*t^2*e^-z))][2t*e^-z]

    ...i plugged in the values for x and y, i'll leave you to simplify this, i have class soon

    similarly, du/dz = du/dx*dx/dz + du/dy*dy/dz

    => du/dz = (ycos(xy))(2e^t) + (xcos(xy))(-t^2*e^-z)
    => du/dz = [t^2*e^-z(cos(2z*e^t*t^2*e^-z)][2e^t] + [2z*e^t(cos(2z*e^t*t^2*e^-z))][-t^2*e^-z]

    ...again, i leave you to simplify this
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