Find the derivative

1. Feb 27, 2005

danne89

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?

2. Feb 27, 2005

neutrino

Redo the first step after the substitution.

3. Feb 27, 2005

danne89

You mean: $\frac{d(1+1/t)}{dt}= - \frac{1}{t^2}$ ??

4. Feb 27, 2005

Pseudo Statistic

Out of curiosity.. why are you doing a substitution? Why don't you just use the quotient rule?

5. Feb 27, 2005

danne89

I dunno.. The Quotient rule is quite nasty, I think.

6. Feb 27, 2005

Pseudo Statistic

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..
$$y = t/(1+t^{-1})$$
$$y' = (1(1+t^{-1})-t(-t^{-2}))/(1+t^{-1})^2$$

Last edited: Feb 27, 2005
7. Feb 27, 2005

dextercioby

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.

8. Feb 27, 2005

neutrino

Nope. I meant $$\frac{d(t(1/u))}{dt} = t\frac{d(1/u)}{dt} + 1/u$$ .

9. Feb 27, 2005

dextercioby

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. Feb 27, 2005

Pseudo Statistic

I wholeheartedly concur with his statement.
K.

11. Feb 27, 2005

danne89

OK. I stick with the Quotient rule...