Derivative of ( (X^3-1)/(X^3+1) )^1/3

[beneakin]

Homework Statement

Find the derivative:

( (X^3-1)/(X^3+1) )^1/3

Homework Equations

d/dx f(g(x)) = f'(g(x)) * g'(x)

quotient rule x/a x'a-xa'/a^2

The Attempt at a Solution

first i used the chain rule and quotient rule to get 1/3 ((x^3-1)/(x^3+1))^-2/3 * ((3x^2(x^3+1) - (x^3-1)3x^2)/(x^3+1)^2)

canceling some thing out on the second part of the neumerator i ended up with

1/3 ((x^3-1)/(x^3+1))^-2/3 * ((6x^2)/(x^3+1)^2)

if i try to simplify more it just get into a hole bunch of nasty fractions... and it just doesn't seem like the right answer, i must have taken a wrong turn some where

thanks a lot!

 

[Rake-MC]

Got the same as you, except on the numerator of the second part it's 6x^2 - 2x^3
like: 1/3 ((x^3-1)/(x^3+1))^-2/3 * ((6x^2 - 2x^3)/(x^3+1)^2)

make sure to get your minus sign right in the numerator when doing the quotient rule.

 

[beneakin]

is that the final answer? it seems a bit large to be?

thanks

 

[Rake-MC]

Is there a rule as to how large answers are allowed to be?

If you really want to get complicated you can simplify it to $$\frac{2x^2}{(x^3-1)^{2/3} (x^3+1)^{4/3}}$$

 

[beneakin]

great thanks so much