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Thread: implicit differentiation

  1. #1
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    implicit differentiation

    (1/xy)=x+y
    find y'
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    $\displaystyle \frac{1}{xy} = x+y $ is the same as $\displaystyle xy(x+y) = 1 $

    $\displaystyle \frac{d}{dx} \left(xy(x+y)\right) = \frac{d}{dx} \left(1 \right) $

    use the product rule

    $\displaystyle xy \frac{d}{dx} \left(x+y \right) + (x+y)\frac{d}{dx} \left(xy \right) = 0 $

    Note: by the chain rule, $\displaystyle \frac{d}{dx} \left(f(y) \right) = f'(y)\frac{dy}{dx} $, so

    $\displaystyle xy(1+\frac{dy}{dx})+(x+y)(y+x\frac{dy}{dx}) = 0 $

    $\displaystyle xy+xy\frac{dy}{dx}+xy+x^{2}\frac{dy}{dx}+y^{2}+xy\ frac{dy}{dx} = 0 $

    $\displaystyle (x^{2}+2xy)\frac{dy}{dx} = -y^{2}-2xy $

    $\displaystyle \frac{dy}{dx} = -\frac{y^{2}+2xy}{x^2+2xy} $
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