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Easy second derivative question

  1. Nov 29, 2007 #1
    [SOLVED] easy second derivative question

    1. The problem statement, all variables and given/known data
    "If y^2 - 3x =7 what is the second derivative?"

    2. Relevant equations

    Answer choices:

    A) -6/7y^3
    B) -3/y^3
    C) 3
    D) 3/2y
    E) -9/4y^3

    3. The attempt at a solution

    I got the first derivative to be: 3/2y
    Second derivative: -6/(4y^2)

    I cannot for the life of me figure out what I did wrong, but I know it's something since that's not a choice. I've been messing up a lot lately, so I think I'm making some sort of vital error. A step by step explanation would be appreciated!

    You guys are so awesome here, thank you for all that you do.
  2. jcsd
  3. Nov 29, 2007 #2


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    Homework Helper

    If y'=3/(2y)=(3/2)*y^(-1) (and it is) then y''=(3/2)*(-1)*y^(-2)*y'. Notice the y' on the right side coming from the chain rule. Now substitute your previous expression for y' for that y'.
  4. Nov 29, 2007 #3


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    well the first derivative is [itex]\frac{dy}{dx}=\frac{3}{2y}=\frac{3}{2}y^{-1}[/itex]
    diff. w.r.t x again


    which should give....[tex]\frac{d^2y}{dx^2}=\frac{-3}{2}y^{-2}\frac{3}{2}y^{-1}[/tex]

    I think you multiplied wrongly
  5. Nov 29, 2007 #4
    Dick, you are beyond wonderful. I kept messing up because I wasn't subbing for y' properly, but now I see the trick to use the first derivative.

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