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Math Help - implicit differentiation

  1. #1
    Super Member bigwave's Avatar
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    Cool implicit differentiation

    find dy/dx of \sqrt{xy} = 1 + x^2y

    rewrite
    (xy)^{1/2}-x^2y = 1
    derivative of terms
    \frac{1}{2}(xy)^{-1/2}(xy' + y) - (x^2y' + y2x) = 0

    \frac{xy'}{2\sqrt{xy}}+\frac{y}{2\sqrt{xy}}-x^2y'-2xy=0

    took more steps but did not get to the answer which is:
    y' = \frac{4xy\sqrt{xy}-y}{x-2x^2\sqrt{xy}}
    Last edited by bigwave; March 10th 2010 at 05:42 PM. Reason: corrected error
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  2. #2
    MHF Contributor
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    Quote Originally Posted by bigwave View Post
    find dy/dx of \sqrt{xy} = 1 + x^2y

    rewrite
    (xy)^{1/2}-x^2y = 1
    derivative of terms
    \frac{1}{2}(xy)^{-1/2}(xy' + y) - (\color{red}x^2\color{black}y' + y2x) = 0

    \frac{xy'}{2\sqrt{xy}}+\frac{y}{2\sqrt{xy}}-\color{red}x^2\color{black}y'-2xy=0

    took more steps but did not get to the answer which is:
    y' = \frac{4xy\sqrt{xy}-y}{x-2x^2\sqrt{xy}}
    hi Bigwave,

    you had an error in an early step.
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  3. #3
    Member Black's Avatar
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    \frac{1}{2}(xy)^{-1/2}(xy'+y)-(x^2y'+2xy)=0

    (x(xy)^{-1/2}-2x^2)y'=4xy-y(xy)^{-1/2}

    y'=\frac{4xy-y(xy)^{-1/2}}{x(xy)^{-1/2}-2x^2}.

    Multiply numerator and denominator by (xy)^1/2 to get the answer.
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  4. #4
    Super Member bigwave's Avatar
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    thanks got it

    thanks got it
    just have to carefull
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