1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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]

    Voilà.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

    User Avatar
    Science Advisor
    Homework Helper

    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...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Find the derivative
Loading...