Question

Question #6ae3e

Answer 1

`f'(x)=4((x^3-1)/(2x^3+1))^3xx(9x^2)/((2x^3-1)^2)`

Explanation:

Assuming that your looking for the derivative of;
`f(x)=[(x^3-1)/(2x^3+1)]^4`
Let `u(x)=(x^3-1)/(2x^3+1)`

`therefore` `f=u^4`

and

`(df)/dx=(df)/(du)*(du)/(dx)` (chain rule)

Using quotient rule to differentiate `u(x)`
REMEMBER - Quotient Rule
`f(x)=u/v`
`f'(x)=(u'v-uv')/v^2`

`u'(x)=((3x^2)(2x^3+1)-(6x^2)(x^3-1))/((2x^3-1)^2)`
`u'(x)=((6x^5+3x^2)-(6x^5-6x^2))/((2x^3-1)^2)`
`u'(x)=(9x^2)/((2x^3-1)^2)`

and, as `f=u^4`, `f'=4u^3`

`therefore` `(df)/dx=(df)/(du)*(du)/(dx)=(4u^3)(9x^2)/((2x^3-1)^2)`
Substituting back in `u(x)=(x^3-1)/(2x^3+1)`
`f'(x)=4((x^3-1)/(2x^3+1))^3xx(9x^2)/((2x^3-1)^2)`

I cant imagine you would have to simplify this in an exam.

hope this helps :)