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Math Help - derivative - using quotient rule

  1. #1
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    derivative - using quotient rule

    how do I get the derivative of:

    -4x
    ---------
    (x^2-1)^2

    the answer is

    12x^2+4
    --------------
    (x^2-1)^3

    using the quotient rule, it seems simple enough but I cant seem to cancel out properly. Thanks!
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  2. #2
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    Just in case a picture helps...



    ... where



    ... is the chain rule, wrapped inside...



    ... the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    The general drift is...



    Similar to the quotient rule, but no automatic squaring of the denominator.
    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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  3. #3
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    Quote Originally Posted by calculus0 View Post
    how do I get the derivative of:

    -4x
    ---------
    (x^2-1)^2

    the answer is

    12x^2+4
    --------------
    (x^2-1)^3

    using the quotient rule, it seems simple enough but I cant seem to cancel out properly. Thanks!
    y = \frac{-4x}{(x^2 - 1)^2}


    \frac{dy}{dx} = \frac{(x^2 - 1)^2\,\frac{d}{dx}(-4x) - (-4x)\,\frac{d}{dx}[(x^2 - 1)^2]}{[(x^2 - 1)^2]^2}

     = \frac{-4(x^2 - 1)^2 + 4x(2x)(2)(x^2 - 1)}{(x^2 - 1)^4}

     = \frac{-4(x^2 - 1)^2 + 16x^2(x^2 - 1)}{(x^2 - 1)^4}

     = \frac{(x^2 - 1)[-4(x^2 - 1) + 16x^2]}{(x^2 - 1)^4}

     = \frac{-4x^2 + 4 + 16x^2}{(x^2 - 1)^3}

     = \frac{12x^2 + 4}{(x^2 - 1)^3}

     = \frac{4(3x^2 + 1)}{(x^2 - 1)^3}.
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  4. #4
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    Next time, show your efforts. Someone will be pleased to help you whereever you wander off.
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  5. #5
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    Dear calculus0,

    By the time I typed the answer it had already been given by other members. Hope these answers will help you.
    Last edited by Sudharaka; March 18th 2010 at 02:16 AM. Reason: Answer already given
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  6. #6
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    THANKS! everyone.. that was so fast.
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