Apparently easy Chain Rule Problem

[rambo5330]

Homework Statement

F(s) = ( s - $$\frac{1}{s^2}$$)³

I have to calculate the derivative of this using chain rule everytime i try i get something way different than in the back of the book... my first move is

3( s - $$\frac{1}{s^2}$$)² X ( 1 + $$\frac{2}{s^3}$$)

is this correct? then expand out from here? maybe there's a problem when i expand.. i don't know but any help would be great thanks...

 

[tiny-tim]

hi rambo5330!

F(s) = ( s - $$\frac{1}{s^2}$$)³

I have to calculate the derivative of this using chain rule everytime i try i get something way different than in the back of the book... my first move is

3( s - $$\frac{1}{s^2}$$)² X ( 1 + $$\frac{2}{s^3}$$)

looks ok to me …

what do you get when you expand it?

 

[Je m'appelle]

Yeah,

$$\frac{d}{ds}F(s) = 3(s - \frac{1}{s^2})^2 (1 + \frac{2}{s^3})$$

Seems fine.

Maybe the author expanded the expression, what answer do you have in the back of the book?

_________________

EDIT: Listen, I've expanded it and what I've found was something like this

$$\frac{d}{ds}F(s) = 3(\frac{s^3-1}{s^2})^2(\frac{s^3 +2}{s^3}) \Rightarrow 3(\frac{(s^3-1)^2}{s^4})(\frac{s^3 +2}{s^3}) = \frac{3}{s^7}((s^3-1)^2(s^3+2))$$

 

[rambo5330]

sorry for late response...

the answer in the text is.

$$\frac{d}{ds}F(s) = \frac{3( s^9 - 3s^3 + 2)}{s^7}$$

when i expand i end up with something similar to yours but i obviously made an error somewhere I'm going to try again right now... i really don't see how they are arriving at this solution

 

[rambo5330]

so i finally arrived at the solution thanks a bunch.. i justt needed to know if i was wrong right off the bat or if it was in my expansion and you jeez.. after awhile of work i found where i made my error.. and i arrived at

$$\frac{d}{ds}F(s) = 3(s^2 - \frac{3}{s^4} + \frac{2}{s^7})$$

which in then became clear that the book cleared the fractions by multiplying/dividing by s⁷

pain in the butt