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Math Help - Need help with 2nd derivative

  1. #1
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    Post Need help with 2nd derivative

    y = t/(t+1)^2

    Finding the first derivative:
    y' = (t+1)^2 (1) - [t(t+1)(2)]/(t+1)^4
    y' = (t+1)[(t+1) - 2t]/(t+1)^4
    y' = (1-t)/(t+1)^3

    Assuming that is correct, then I'll move on to the second derivative:
    y" = (t+1)^3(-1) - [(1-t)(3)(t+1)^2]/(t+1)^6
    y" = -(t+1)^2[3(1-t)]/(t+1)^6 I really start having problems here:
    y" = (1-t)^2[(3-3t)/(t+1)^6]

    Hmmm....could you give me a boost?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by becky View Post
    y = t/(t+1)^2

    Finding the first derivative:
    y' = (t+1)^2 (1) - [t(t+1)(2)]/(t+1)^4
    y' = (t+1)[(t+1) - 2t]/(t+1)^4
    y' = (1-t)/(t+1)^3

    Assuming that is correct, then I'll move on to the second derivative:
    y" = (t+1)^3(-1) - [(1-t)(3)(t+1)^2]/(t+1)^6
    y" = -(t+1)^2[3(1-t)]/(t+1)^6 I really start having problems here:
    y" = (1-t)^2[(3-3t)/(t+1)^6]

    Hmmm....could you give me a boost?
    First, you really need to put () around the numerator in the first line!

    You factored the second line wrong.

    y" = {(t+1)^3(-1) - [(1-t)(3)(t+1)^2]}/(t+1)^6

    Taking a common (t+1)^2 from the first and second terms gives:
    y'' = (t+1)^2{-(t+1) - 3(1-t)}/(t+1)^6

    y'' = (t+1)^2{2t-4}/(t+1)^6

    y'' = 2(t-2)/(t+1)^4

    -Dan
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