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

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The discussion centers on finding the derivative of the function y=t/(1+1/t). Participants debate the use of substitution versus the Quotient rule for differentiation, with many arguing that the Quotient rule is more straightforward despite its complexity. One user demonstrates the correct application of the Quotient rule, yielding a simpler and more effective solution. There is consensus that the substitution method complicates the process unnecessarily. Ultimately, the group agrees that sticking with the Quotient rule is the best approach for this problem.
danne89
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Find the derivative,
y=\frac{t}{1+1/t} = t * 1/u
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}

What have I done wrong?
 
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Redo the first step after the substitution.
 
You mean: \frac{d(1+1/t)}{dt}= - \frac{1}{t^2} ??
 
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..
y = t/(1+t^{-1})
y' = (1(1+t^{-1})-t(-t^{-2}))/(1+t^{-1})^2 :approve:
 
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It's pretty useless to use a substitution,if u're not doing it properly.Better use the Quotient rule:
\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}}

Voilà.Piece of cake.

Daniel.
 
danne89 said:
You mean: \frac{d(1+1/t)}{dt}= - \frac{1}{t^2} ??
Nope. I meant \frac{d(t(1/u))}{dt} = t\frac{d(1/u)}{dt} + 1/u .
 
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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