Finding the derivative of y=t/(1+1/t)

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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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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:
 
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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]

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...
 
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.