Find Derivative of y=t/(1+1/t)

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In summary, Daniel is trying to find the derivative of y=t/(1+t^{-1}) but gets confused and uses a substitution instead. K agrees that substitutions are okay for integrals, but finds the Quotient rule easier to understand.
  • #1
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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  • #2
Redo the first step after the substitution.
 
  • #3
You mean: [itex]\frac{d(1+1/t)}{dt}= - \frac{1}{t^2}[/itex] ??
 
  • #4
Out of curiosity.. why are you doing a substitution? Why don't you just use the quotient rule?
 
  • #5
I dunno.. The Quotient rule is quite nasty, I think.
 
  • #6
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:
  • #7
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.
 
  • #8
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] .
 
  • #9
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...
 

1. What is the derivative of y=t/(1+1/t)?

The derivative of y=t/(1+1/t) is given by the following formula:
dy/dt = (1+1/t)^(-2) * (1/t^2)

2. How do you find the derivative of y=t/(1+1/t)?

To find the derivative of y=t/(1+1/t), we can use the quotient rule for differentiation. This rule states that the derivative of a quotient is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

3. Can the derivative of y=t/(1+1/t) be simplified?

Yes, the derivative of y=t/(1+1/t) can be simplified by expanding the expression and combining like terms. The final simplified form is dy/dt = (1+2/t) / (t^2 * (1+1/t)^2).

4. What is the significance of the derivative of y=t/(1+1/t)?

The derivative of y=t/(1+1/t) represents the rate of change of the function at any given point. It tells us how fast the function is changing at that point and in which direction.

5. How can the derivative of y=t/(1+1/t) be applied in real-life scenarios?

The derivative of y=t/(1+1/t) can be applied in various real-life scenarios, such as in physics to calculate the velocity of an object, in economics to determine the marginal cost of production, and in engineering to optimize the performance of a system. It is also used in calculus to find the maximum and minimum values of a function.

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