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Math Help - implicit differentiation

  1. #1
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    implicit differentiation

    Problem: x^2+xy-y^2=4

    Solution attempt:  \frac{d}{dx} (x^2)+\frac{d}{dx}(xy)-\frac{d}{dx}(y^2)=\frac{d}{dx}(4)

    2x+y+xy'-2yy'=0

    (2x+y)+(xy'-2yy')=0

    y'(x-2y)=-(2x+y)

    y'=\frac{-(2x+y)}{x-2y}

    y'=\frac{-2x-y}{-2y+x}

    y'=\frac{2x-y}{2y+x}

    Answer in book: y'=\frac{2x+y}{2y-x}

    Notes:I've always had a mundane problem with negatives and fractions, when I cancel out the negs in the last step, must I also switch the signs of the proceeding variables as well?
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  2. #2
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    Your second to last step is correct. What you are doing when you "cancel out" the negatives is multiplying the fraction by \frac{-1}{-1}. So you get \frac{-1}{-1} \cdot \frac{-2x - y}{-2y + x} = \frac{(-1)(-2x - y)}{(-1)(-2y + x)}. Then use the distributive property to arrive at the correct answer (the one in your book).
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  3. #3
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     \frac{d}{dx} (x^2)+\frac{d}{dx}(xy)-\frac{d}{dx}(y^2)=\frac{d}{dx}(4)

    2x+y+xy'-2yy'=0

    (2x+y)=2yy'-xy'

    (2x+y)=y'(2y-x)

    y'=\frac{2x+y}{2y-x}


    which matches your answer: y'=\frac{2x+y}{2y-x}
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  4. #4
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    case closed, thank you both
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