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Math Help - Differentiating Implicitly

  1. #1
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    Differentiating Implicitly

    Is it possible to differentiate this equation?

    (2x^2)(y)-(7y^2)=(8x)-2y-11 to find y'(1)?
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  2. #2
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    2yx^2 - 7y^2 - 8x + 2y = -11

    -(d/dx)/(d/dy)

    (8 - 4xy) / (2x^2 - 14y + 2)

    y'(1) just accounts for one of those variables, however. You need an x and a y coordinate to perform the calculation. For y'(1), do you mean x by that 1? If so, just plug back into the original equation to get the y value accompanying it.

    Edit: Missed that 2y', sorry. Fixed
    Last edited by derfleurer; April 20th 2009 at 11:17 AM.
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  3. #3
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    Quote Originally Posted by derfleurer View Post
    2yx^2 - 7y^2 - 8x + 2y = -11

    -(d/dx)/(d/dy)

    (8 - 4xy) / (2x^2 - 14y)

    y'(1) just accounts for one of those variables, however. You need an x and a y coordinate to perform the calculation.
    I was wondering why I couldn't do it either. Thank you.
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  4. #4
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    Hello, Hakhengkim!

    Did you follow derfleurer's advice and finish the problem?


    2x^2y-7y^2\:=\:8x-2y-11. . Find y'(1)
    Differentiate implicitly . . .

    . . 2x\,\frac{dy}{dx} + 4xy - 14y\,\frac{dy}{dx} \:=\:8 - 2\,\frac{dy}{dx} \quad\Rightarrow\quad 2x\,\frac{dy}{dx} - 14y\,\frac{dy}{dx} + 2\,\frac{dy}{dx} \:=\:8 - 4xy

    Factor: . 2(x-7y - 6)\,\frac{dy}{dx} \:=\:4(2-xy) \quad\Rightarrow\quad \frac{dy}{dx} \;=\;\frac{2(2-xy)}{x-7y-6}


    If x = 1, the equation becomes: . 2y - 7y^2 \:=\:8 - 2y - 11 \quad\Rightarrow\quad 7y^2 - 4y - 3 \:=\:0

    . . (y - 1)(7y+3) \:=\:0 \quad\Rightarrow\quad y \:=\:1,\:-\tfrac{3}{7}

    And we have two points: . (1,\,1),\;\;\left(1,\,-\tfrac{3}{7}\right)


    At (1,\,1)\!:\;\;\frac{dt}{dx} \:=\:\frac{2(2-1)}{1-7-6} \:=\:-\frac{1}{6}

    At \left(1,\,-\tfrac{3}{7}\right)\!:\;\;\frac{dy}{dx} \:=\:\frac{2\left[2-(1)(-\frac{3}{7})\right]}{1 - 7(-\frac{3}{7}) - 6} \:=\:-\frac{17}{7}

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  5. #5
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    Whoop, missed that + 2 in the denominator.

    Not to hijack the thread, but how'd you get 2xy' - 14yy' +2y' = 2(x - 7y - 6)y'?
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