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Math Help - Differentiation! Please help!

  1. #1
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    Question Differentiation! Please help!

    I'm working on Survey of Calculus.
    Please help!

    Evaluate dy/dt for the function at the point.

    x^3 e^y - y^3 ln x = 7; dx/dt = 2, x = 1, y = 3
    Last edited by Matho; November 28th 2008 at 03:46 PM.
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  2. #2
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    Hello, Matho!

    Evaluate \frac{dy}{dt} for the function at the point.
    . . . x^3e^y - y^3\ln x \:=\: 7,\quad x=1,\;y=3,\;\frac{dx}{dt} = 2
    Differentiate implicitly ... (Don't forget the Product Rule!)

    . . x^3e^y\,\frac{dy}{dt} + 3x^2e^y\,\frac{dx}{dt} - y^3\,\frac{1}{x}\,\frac{dx}{dt} - 3y^2\ln x\,\frac{dy}{dt} \;=\;0


    Substitute the given values:

    . . (1^3)(e^3)\,\frac{dy}{dt} + (3)(1^2)(e^3)(2) - (3^3)\left(\tfrac{1}{1}\right)(2) - (3)(3^2)(\ln 1)\,\frac{dy}{dt} \;=\;0

    . . e^3\!\cdot\!\frac{dy}{dt} + 6e^3 - 54 - 27\!\cdot\!0\!\cdot\!\frac{dy}{dt} \;=\;0

    . . e^3\!\cdot\!\frac{dy}{dt} \;=\;54 - 6e^3

    . . \frac{dy}{dt} \;=\;\frac{6(9-e^3)}{e^3}

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