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Is this correct take the derivative of

  1. Apr 11, 2005 #1
    (t-1)^1/2*(t^-2)

    I hope you guys can understand what im trying to say up there.... SO i did the product rule, and my answer was this.. lemmi know if it's correct...thanks a bunch.

    (t-1)^-1/2 t^-3 { -3/2t +2 }

    That was my answer...
     
  2. jcsd
  3. Apr 11, 2005 #2

    dextercioby

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    I get

    [tex] \frac{1}{2}\left[(t-1)^{-\frac{1}{2}}\right] t^{-3} \left(4-3t\right)[/tex]

    Which is the same thing,so everything is okay. :smile:

    Daniel.

    P.S.Lakers missed the play-offs :wink:
     
    Last edited: Apr 11, 2005
  4. Apr 11, 2005 #3
    hey that looks perfect to me!
     
  5. Apr 11, 2005 #4
    is this what you trying to differentiate - [tex] (t-1)^{\frac{t^2}{2}} [/tex] or is this
    [tex] (t-1)^{\frac{1}{2}} \frac{t^2}{2} [/tex]
     
  6. Apr 11, 2005 #5

    dextercioby

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    Nope.

    [tex] (t-1)^{\frac{1}{2}} t^{-2} [/tex]

    Daniel.
     
  7. Apr 11, 2005 #6
    I can see where the [tex] \frac{1}{2}\left[(t-1)^{-\frac{1}{2}}\right][/tex] came from but the [tex]t^{-3} \left(4-3t\right)[/tex] has lost me. What did you do to get that because I would have just done [tex]-2t^{-3}[/tex]

    The Bob (2004 ©)
     
  8. Apr 11, 2005 #7

    dextercioby

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    I forced something as a factor (v.above),and that's how i ended up with that paranthesis.

    Daniel.
     
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