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Math Help - 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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     \frac{1}{xy} = x+y is the same as  xy(x+y) = 1

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

    use the product rule

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

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

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

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

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

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