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Math Help - Chain rule derivative- double check this please?

  1. #1
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    [SOLVED] Chain rule derivative- double check this please?

    I'm not sure if I did this correctly.

    s(t)= √(t^2 + 2t)
    s'(t)= (t + 1)/[√(t^2 + 2t)]
    Here's where I have trouble:
    s''(t)= (t^2 + 2t)^(-1/2) (t + 1)
    = (-1/2)(t^2 + 2t)^(-3/2) (t + 1) + (t^2 + 2t)^(-1/2) (1)
    = 1/[√(t^2 + 2t)] - (t+1)/[2(t^2 + 2t)^(3/2)]
    Is this the correct answer?
    Last edited by Goose; November 10th 2010 at 11:19 AM.
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  2. #2
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  3. #3
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    Thanks, didn't know about that site...
    s''(t)
    So I'm not right, but close! The (t + 1)(2t + 2) numerator isn't making sense to me...
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  4. #4
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    Ah! Perhaps if I use the chain rule correctly!
    s''(t)= (t^2 + 2t)^(-1/2) (t + 1)
    = (-1/2)(t^2 + 2t)^(-3/2) (2t + 2)(t + 1) + (t^2 + 2t)^(-1/2) (1)
    = 1/[√(t^2 + 2t)] - (t+1)(2t+2)/[2(t^2 + 2t)^(3/2)]

    I hate when I make mistakes like that.
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  5. #5
    A Plied Mathematician
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    You're welcome for the link. The site basically uses Mathematica syntax.
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