Simplifying problem

[cosmic]

Hi guys,

I'm having difficulties understanding this. Any help would be appreciated.

Let's say I have a function f(t)=(t^3*(1/(t+1)) - ln(t+1)*3t^2)/t^6 I know this simplifies to f(t)=(t/(1+t)-3 ln(1+t))/t^4. What I'm trying to understand is how the t^3 simplifies to t. I think I'm correct in thinking that the 3t^2 cancels out with the t^6 to leave t^4 in the denominator and just 3 in the numerator? What am I missing when it comes to the t^3? Any help would be great.

Thanks in advance.

 

[cosmic]

Nevermind guys, I figured it out. I just didn't factor it out correctly at the start.

Live and learn I guess.

 

[lookagain]

Let's say I have a function f(t)=(t^3*(1/(t+1)) - ln(t+1)*3t^2)/t^6

I know this simplifies to f(t)=(t/(1+t)-3 ln(1+t))/t^4. What I'm trying to understand is how the t^3 simplifies to t.

I think I'm correct in thinking that the 3t^2 cancels out with the t^6 to leave t^4 in the denominator and just 3
in the numerator?

What am I missing when it comes to the t^3?

Think of this approach:


\(\displaystyle f(t) \ = \ \dfrac{t^3\bigg(\dfrac{1}{t + 1}\bigg) \ - \ ln(t + 1)*3t^2}{t^6}\)



\(\displaystyle f(t) \ = \ \dfrac{t^2\bigg[t\bigg(\dfrac{1}{t + 1}\bigg) \ - \ ln(t + 1)*3\bigg]}{t^2(t^4)}\)



\(\displaystyle f(t) \ = \ \dfrac{t\bigg(\dfrac{1}{t + 1}\bigg) \ - \ ln(t + 1)*3}{t^4}\)



\(\displaystyle f(t) \ = \ \dfrac{\dfrac{t}{t + 1} \ - \ 3* ln(t + 1)}{t^4} \ \ \ \ \ \ \ \ \ \)This is the form you mentioned, but it contains a fraction within the larger fraction.
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)For me, I would multiply by the least common denominator, t + 1, to get:


\(\displaystyle f(t) \ = \ \dfrac{t \ - \ 3(t + 1)ln(t + 1)}{(t + 1)t^4}\)

 

[cosmic]

Think of this approach:


\(\displaystyle f(t) \ = \ \dfrac{t^3\bigg(\dfrac{1}{t + 1}\bigg) \ - \ ln(t + 1)*3t^2}{t^6}\)



\(\displaystyle f(t) \ = \ \dfrac{t^2\bigg[t\bigg(\dfrac{1}{t + 1}\bigg) \ - \ ln(t + 1)*3\bigg]}{t^2(t^4)}\)



\(\displaystyle f(t) \ = \ \dfrac{t\bigg(\dfrac{1}{t + 1}\bigg) \ - \ ln(t + 1)*3}{t^4}\)



\(\displaystyle f(t) \ = \ \dfrac{\dfrac{t}{t + 1} \ - \ 3* ln(t + 1)}{t^4} \ \ \ \ \ \ \ \ \ \)This is the form you mentioned, but it contains a fraction within the larger fraction.
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)For me, I would multiply by the least common denominator, t + 1, to get:


\(\displaystyle f(t) \ = \ \dfrac{t \ - \ 3(t + 1)ln(t + 1)}{(t + 1)t^4}\)

Thanks that's great.

Just out of interest when working with derivatives how far do you have to simplify the answer for it to be deemed acceptable?

 

[Quaid]

how far do you have to simplify the answer for it to be deemed acceptable?

Acceptable to whom?

Ask that person!

 

[cosmic]

Acceptable to whom?

Ask that person!

Just meant in general Quaid. Most derivatives I've looked at are simplified to a greater or a lesser degree in various textbooks so I was wondering how far are you expected to simplify a derivative in general.

 

[Quaid]

meant in general Quaid

how far are you expected to simplify

There is no general expectation, cosmic.

Here's an example, taken from a lecture-video at the University of Ohio. The exercise asked students to use the Quotient Rule. The professor wrote out what he considers an acceptable result; this answer is correct, yet it's not simplified at all.



There is no global standard to which all instructors must adhere; different classrooms can have different expectations. If this is an issue for you, ask your instructor for clarification.

Ciao :cool:

 

[cosmic]

There is no general expectation, cosmic.

Here's an example, taken from a lecture-video at the University of Ohio. The exercise asked students to use the Quotient Rule. The professor wrote out what he considers an acceptable result; this answer is correct, yet it's not simplified at all.

View attachment 4211

There is no global standard to which all instructors must adhere; different classrooms can have different expectations. If this is an issue for you, ask your instructor for clarification.

Ciao :cool:

Thanks Quaid.

The textbook I was looking at mentioned that you should simplify the derivatives but not to what extend but I'll ask my tutor.