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Math Help - implicit differentiation- where did i make a mistake

  1. #1
    Junior Member LexiRae's Avatar
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    implicit differentiation- where did i make a mistake

    find d^2y/dx^2 at (1,1) for:
    x^2y + x^2 -y^2 = 1

    i worked it all out but the answer im getting is not an option. could someone show me where i made a mistake? im not asking you to do the problem for me, just fix my error. this is what i did:
    first i found dy/dx:

    x^2(dy/dx) + y(2x) +2x -2y(dy/dx)=0
    (x^2 - 2y)(dy/dx) = -2xy - 2x
    dy/dx = (-2xy-2x)/(x^2-2y)
    when i put (1,1) in, i got 4

    then to find the second derivative i used the quotient rule:
    d^2y/dy^2 = [(x^2-2y) (-2x)(dy/dx) +y(-2) -2] - ((-2xy-2x)(2x-2)(dy/dx)]/(x^2-2y)^2

    then i just put the values in. for the x's and y's i put in 1, and for dy/dx i put in 4. i ended up with:
    [(-1 x -2 x 4 + -2 -2) - (-2 -2) x (2-2(4) ] / 1 = 28 ? but this answer is incorrect. what did i do wrong?


    Thanks so much!
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by LexiRae View Post
    find d^2y/dx^2 at (1,1) for:
    x^2y + x^2 -y^2 = 1

    i worked it all out but the answer im getting is not an option. could someone show me where i made a mistake? im not asking you to do the problem for me, just fix my error. this is what i did:
    first i found dy/dx:

    x^2(dy/dx) + y(2x) +2x -2y(dy/dx)=0
    (x^2 - 2y)(dy/dx) = -2xy - 2x
    dy/dx = (-2xy-2x)/(x^2-2y)
    when i put (1,1) in, i got 4

    then to find the second derivative i used the quotient rule:
    d^2y/dy^2 = [(x^2-2y) (-2x)(dy/dx) +y(-2) -2] - ((-2xy-2x)(2x-2)(dy/dx)]/(x^2-2y)^2

    then i just put the values in. for the x's and y's i put in 1, and for dy/dx i put in 4. i ended up with:
    [(-1 x -2 x 4 + -2 -2) - (-2 -2) x (2-2(4) ] / 1 = 28 ? but this answer is incorrect. what did i do wrong?


    Thanks so much!
    I'm struggling to read all that. Your expression for dy/dx is correct and the value at (1, 1) is correct. I suggest multiplying the top and bottom by -1 so that

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

    Then \frac{d^2y}{dx^2} = \frac{\left(2y + 2x \frac{dy}{dx} + 2\right) (2y - x^2) - \left(2 \frac{dy}{dx} - 2x \right) (2xy + 2x)}{(2y - x^2)^2}.

    Now substitute x = 1, y = 1 and \frac{dy}{dx} = 4. I get -12.
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