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Math Help - implicit differentiation...02

  1. #1
    Junior Member
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    implicit differentiation...02

    problem: 1+x = sin(xy^2)

    solution attempt: differentiating both sides, I have
     1 = -cosx \cdot\frac{d}{dx}[xy^2]

    1 = -cosx(y^2+2xyy')

    dividing by -cosx on both sides, I have
    -\frac{1}{cosx} =y^2+2xyy'

    dividing by 2xy on both sides, I have
     -\frac{2xy}{cosx} = \frac{y^2}{2xy} + y'

    subtracting \frac{y^2}{2xy} from both sides, I have
     y' = -\frac{2xy}{cosx}-\frac{y^2}{2xy}

    then,
     y' = -\frac{y^2(4x^2 - cosx)}{2xycosx}
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  2. #2
    Super Member
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    You differentiated the right side wrong.

    \frac{d}{dx}sin(xy^2)=cos(xy^2)\frac{d}{dx}(xy^2)=  (y^2+2xyy')cos(xy^2)

    Remember when you use the chain rule you have to keep whatever is in the parenthesis.
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