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Math Help - Another Implicit Diff Problem

  1. #1
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    Another Implicit Diff Problem

    Solving for y'

    f(x) = 8x^{2} + 3xy - y^{2} = 2

    16x + (y)(3)y' + (3x)(1)y' - 2yy' = 0 - Product Rule on 3xy

    16x + (3)(y)y' + (3x)(1)y' - 2yy' = 0

    16x + 3yy' + 3xy' - 2yy' = 0

    16x + y'(3y + 3x - 2y) = 0

    y'(3y + 3x - 2y) = -16x

    y' = \dfrac{16x}{3y + 3x - 2y}
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  2. #2
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    Re: Another Implicit Diff Problem

    Hello, Jason76!

    Your "product rule" is incorrect.


    f(x) \:=\:8x^2 + 3xy - y^2 \:=\:0\quad \text{ Find }y'.

    We have: . . . . 8x^2 + 3xy - y^2 \:=\:0

    Then: . 16x + \overbrace{3xy' + 3y} - 2yy' \:=\:0

    n . . . . . . . . . . . . 3xy' - 2yy' \:=\:-16x - 3y

    n . . . . . . . . . . . . (3x-2y)y' \:=\:-(16x+3y)

    . . . . . . . . . . . . . . . . . . . . y' \:=\:-\frac{16x+3y}{3x-2y}
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  3. #3
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    Re: Another Implicit Diff Problem

    How about this?

    Right??

    f(x) = 8x^{2} + 3xy - y^{2} = 2

    16x + (y)(3) + (3x)(1)y' - 2yy' = 0 - Product Rule on 3xy

    16x + (3)(y) + (3x)(1)y' - 2yy' = 0

    16x + 3y + 3xy' - 2yy' = 0

    16x + 3y + y'(3x - 2y) = 0

    y'(3x - 2y) = -16x - 3y

    y'= \dfrac{-16x - 3y}{3y - 2y}
    Last edited by Jason76; September 27th 2013 at 11:30 AM.
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