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Math Help - derivative

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    derivative

    trying to differentiate:
    1-x^2 / (x^2+1)^2

    using the quotient rule,

    u = (1-x^2)
    du/dx = -2x

    v = (x^2+1)^2)
    dv/dx = 4x(1+x^2)

    [u * dv/dx ] - [v *du/dx] / v^2


    I've been given an answer to complete working out to:
    2x(x^2-3)/(x^2+1)^3


    I've managed to work it out up to
    2x(x^2-1)/(x^2+1)^3

    but have no idea where the 3 in the the provided answer came from
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  2. #2
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    Quote Originally Posted by JohnnyB View Post
    trying to differentiate:
    1-x^2 / (x^2+1)^2

    using the quotient rule,

    u = (1-x^2)
    du/dx = -2x

    v = (x^2+1)^2)
    dv/dx = 4x(1+x^2)

    [u * dv/dx ] - [v *du/dx] / v^2


    I've been given an answer to complete working out to:
    2x(x^2-3)/(x^2+1)^3


    I've managed to work it out up to
    2x(x^2-1)/(x^2+1)^3

    but have no idea where the 3 in the the provided answer came from
    \frac{d}{dx} \left[\frac{1-x^2}{(x^2+1)^2}\right]

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

    \frac{(x^2+1) \cdot (-2x) - (1-x^2) \cdot 4x}{(x^2+1)^3}<br />

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

    \frac{2x^3 - 6x}{(x^2+1)^3}

    \frac{2x(x^2-3)}{(x^2+1)^3}
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  3. #3
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    Joined
    Mar 2010
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    uuughhhhhh, so easy.

    made a typo in my working out.

    *slaps head*
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