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Homework Help: Find the derivative

  1. Feb 27, 2005 #1
    Find the derivative,
    y=[itex]\frac{t}{1+1/t} = t * 1/u[/itex]
    [tex]y'= \frac{d(1/u)}{dt} + 1/u = - \frac{1}{u^2}* \frac{du}{dt} + 1/u = - \frac{1}{(1+1/t)^2}* \frac{d(1+1/t)}{dt} + \frac{1}{1+1/t} = - \frac{1}{(1+1/t)^2} * ( - \frac{1}{t^2}) + \frac{1}{1+1/t} = \frac{1}{(1+1/t)^2 * t^2} + \frac {1}{1+1/t}[/tex]

    What have I done wrong?
     
  2. jcsd
  3. Feb 27, 2005 #2
    Redo the first step after the substitution.
     
  4. Feb 27, 2005 #3
    You mean: [itex]\frac{d(1+1/t)}{dt}= - \frac{1}{t^2}[/itex] ??
     
  5. Feb 27, 2005 #4
    Out of curiosity.. why are you doing a substitution? Why don't you just use the quotient rule?
     
  6. Feb 27, 2005 #5
    I dunno.. The Quotient rule is quite nasty, I think.
     
  7. Feb 27, 2005 #6
    Yeah, it's very nasty... but it's kinda easier if you try it... (In the end it looks quite sexy, too :D)

    To make this post more productive, you could as well start out with..
    [tex]y = t/(1+t^{-1})[/tex]
    [tex]y' = (1(1+t^{-1})-t(-t^{-2}))/(1+t^{-1})^2[/tex] :approve:
     
    Last edited: Feb 27, 2005
  8. Feb 27, 2005 #7

    dextercioby

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    It's pretty useless to use a substitution,if u're not doing it properly.Better use the Quotient rule:
    [tex] \frac{d}{dt}(\frac{t}{1+\frac{1}{t}})=\frac{d}{dt}(\frac{t^{2}}{t+1})=\frac{2t(t+1)-t^{2}}{(t+1)^{2}}=\frac{t^{2}+2t}{(t+1)^{2}} [/tex]

    VoilĂ .Piece of cake.

    Daniel.
     
  9. Feb 27, 2005 #8
    Nope. I meant [tex]\frac{d(t(1/u))}{dt} = t\frac{d(1/u)}{dt} + 1/u[/tex] .
     
  10. Feb 27, 2005 #9

    dextercioby

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    I simply do not understand why you keep insisting on this method,which is very unintuitive and,as it has been easily proven by your posts,brings a lot complication than it was supposed to.

    Daniel.

    P.S.Substitutions are okay for integrals...
     
  11. Feb 27, 2005 #10
    I wholeheartedly concur with his statement.
    K.
     
  12. Feb 27, 2005 #11
    OK. I stick with the Quotient rule...
     
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