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Math Help - Implicit differentiation (2 different answers!)

  1. #1
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    Implicit differentiation (2 different answers!)

    Trying to find dy/dx for x^2/(x-y^2)=5

    Method 1: x^2 = 5x - 5y^2
    2x = 5 - 10y(dy/dx)
    dy/dx = (5-2x)/10y

    Method 2: Quotient rule
    2x(x-y^2) - x^2(1-2y(dy/dx)) all over (x-y^2)^2 =0
    2x(x-y^2) = x^2(1-2y(dy/dx))
    2(x-y^2)/x = 1 - 2y (dy/dx)
    2y(dy/dx) = 1 - 2(x-y^2)/x = 2(y^2-x)/x + 1 = (2y^2 - x)/x
    dy/dx = (2y^2 - x)/2xy

    Two different answers, not algebraically equivalent. Where have I gone wrong? Which one is right?
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  2. #2
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    Quote Originally Posted by pingleby View Post
    Trying to find dy/dx for x^2/(x-y^2)=5

    Method 1: x^2 = 5x - 5y^2
    2x = 5 - 10y(dy/dx)
    dy/dx = (5-2x)/10y

    Method 2: Quotient rule
    2x(x-y^2) - x^2(1-2y(dy/dx)) all over (x-y^2)^2 =0
    2x(x-y^2) = x^2(1-2y(dy/dx))
    2(x-y^2)/x = 1 - 2y (dy/dx)
    2y(dy/dx) = 1 - 2(x-y^2)/x = 2(y^2-x)/x + 1 = (2y^2 - x)/x
    dy/dx = (2y^2 - x)/2xy

    Two different answers, not algebraically equivalent. Where have I gone wrong? Which one is right?
    These answers are algebraically equivalent! Use the initial equation from Method 1 (x^2 = 5x - 5y^2) to make the substitution y^2 = x(1(x/5)) in the second answer, and see it reduce to the first answer.
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  3. #3
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    ^^ Like he said ^^

    Here's how I would do it:

    \frac{x^2}{x-y^2} = 5

    x^2 = 5x - 5y^2

    2x = 5 - 10y\tfrac{dy}{dx}

    \tfrac{dy}{dx} = \frac{2x - 5}{-10y} = \boxed{\frac{5-2x}{10y}}

    Which is your method 1.
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