Where Did I Go Wrong in Finding the Derivative of y=t/(1+1/t)?

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Homework Help Overview

The discussion revolves around finding the derivative of the function y = t/(1 + 1/t). Participants are examining different methods for differentiation, particularly focusing on the use of substitution versus the quotient rule.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the original poster's substitution method and question its effectiveness. Some suggest using the quotient rule instead, while others express concerns about the complexity of the quotient rule. There are also clarifications sought regarding the derivative of specific components of the function.

Discussion Status

The discussion is ongoing, with various methods being debated. Some participants have offered guidance on using the quotient rule, while others are questioning the necessity and effectiveness of the substitution approach. There is no explicit consensus on the best method to use.

Contextual Notes

Participants note that the original poster's approach may not be yielding the desired clarity, leading to complications in the differentiation process. There is an acknowledgment that substitutions can be more appropriate in certain contexts, such as integrals.

danne89
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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?
 
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Redo the first step after the substitution.
 
You mean: [itex]\frac{d(1+1/t)}{dt}= - \frac{1}{t^2}[/itex] ??
 
Out of curiosity.. why are you doing a substitution? Why don't you just use the quotient rule?
 
I dunno.. The Quotient rule is quite nasty, I think.
 
danne89 said:
I dunno.. The Quotient rule is quite nasty, I think.
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:
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.
 
danne89 said:
You mean: [itex]\frac{d(1+1/t)}{dt}= - \frac{1}{t^2}[/itex] ??
Nope. I meant [tex]\frac{d(t(1/u))}{dt} = t\frac{d(1/u)}{dt} + 1/u[/tex] .
 
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...
 
  • #10
dextercioby said:
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...
I wholeheartedly concur with his statement.
K.
 
  • #11
OK. I stick with the Quotient rule...
 

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