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Math Help - Differentiate

  1. #1
    Member Awsom Guy's Avatar
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    Differentiate

    Check this derivation please.

    s= 3t/(t^3-1)

    i have worked so far into this:

    (3t)(t^3-1)^-1
    (-3t)(t^3-1)^-2
    (-3t)/(t^3-1)^2

    I am not sure if I am doing this right.
    Could some one check so far and finish it off for me.
    The answer should be: -3(1+2t^3)/(t^3-1)^2.
    I am not sure where the top (1+2t^3) comes from please check.

    p.s = sorry about the quick posts i've done thanks for the help I have recieved. Thanks
    Last edited by Awsom Guy; February 8th 2010 at 01:23 AM. Reason: not derivative but Differentiation
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by Awsom Guy View Post
    Check this derivation please.

    s= 3t/(t^3-1)

    i have worked so far into this:

    (3t)(t^3-1)^-1
    (-3t)(t^3-1)^-2
    (-3t)/(t^3-1)^2

    I am not sure if I am doing this right.
    Could some one check so far and finish it off for me.
    The answer should be: -3(1+2t^3)/(t^3-1)^2.
    I am not sure where the top (1+2t^3) comes from please check.

    p.s = sorry about the quick posts i've done thanks for the help I have recieved. Thanks
    Have you learnt the quotient rule?

    Check it yourself here: differentiate 3t/(t^3-1) - Wolfram|Alpha

    Click on Show steps.
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  3. #3
    MHF Contributor
    Prove It's Avatar
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    Quote Originally Posted by Awsom Guy View Post
    Check this derivation please.

    s= 3t/(t^3-1)

    i have worked so far into this:

    (3t)(t^3-1)^-1
    (-3t)(t^3-1)^-2
    (-3t)/(t^3-1)^2

    I am not sure if I am doing this right.
    Could some one check so far and finish it off for me.
    The answer should be: -3(1+2t^3)/(t^3-1)^2.
    I am not sure where the top (1+2t^3) comes from please check.

    p.s = sorry about the quick posts i've done thanks for the help I have recieved. Thanks
    Use the quotient rule:

    s = \frac{3t}{t^3 - 1}


    \frac{ds}{dt} = \frac{(t^3 - 1)\frac{d}{dt}(3t) - 3t\frac{d}{dt}(t^3 - 1)}{(t^3 - 1)^2}

     = \frac{3(t^3 - 1) - 3t(3t^2)}{(t^3 - 1)^2}

     = \frac{3t^3 - 3 - 9t^3}{(t^3 - 1)^2}

     = \frac{-6t^3 - 3}{(t^3 - 1)^2}

     = \frac{-3(2t^3 + 1)}{(t^3 - 1)^2}.
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  4. #4
    Member Awsom Guy's Avatar
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    oh my,,, I am so sorry about that I have forgotten my old work thank you for reminding me . I have correctly done the question thanks.
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  5. #5
    Member Awsom Guy's Avatar
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    Thanks you again.

    Quote Originally Posted by Prove It View Post
    Use the quotient rule:

    s = \frac{3t}{t^3 - 1}


    \frac{ds}{dt} = \frac{(t^3 - 1)\frac{d}{dt}(3t) - 3t\frac{d}{dt}(t^3 - 1)}{(t^3 - 1)^2}

     = \frac{3(t^3 - 1) - 3t(3t^2)}{(t^3 - 1)^2}

     = \frac{3t^3 - 3 - 9t^3}{(t^3 - 1)^2}

     = \frac{-6t^3 - 3}{(t^3 - 1)^2}

     = \frac{-3(2t^3 + 1)}{(t^3 - 1)^2}.


    Thanks again but I figured it out sorry for troubling you
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  6. #6
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    Quote Originally Posted by Awsom Guy View Post
    Check this derivation please.

    s= 3t/(t^3-1)

    i have worked so far into this:

    (3t)(t^3-1)^-1
    You can do it this way, using the product rule rather than the quotient rule- but you didn't use the product rule.
    (fg)'= f'g+ fg'
    (3t(t^3- 1)^{-1})'= (3t)'(t^3-1)^{-1}+ (3t)((t^3-1)^{-1})'
     (3)(t^-1)^{-1}+ (3t)(-1(t^3- 1)^{-2}(3t^2)

    (-3t)(t^3-1)^-2
    (-3t)/(t^3-1)^2

    I am not sure if I am doing this right.
    Could some one check so far and finish it off for me.
    The answer should be: -3(1+2t^3)/(t^3-1)^2.
    I am not sure where the top (1+2t^3) comes from please check.

    p.s = sorry about the quick posts i've done thanks for the help I have recieved. Thanks
    Follow Math Help Forum on Facebook and Google+

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