Question #6ae3e

1 Answer
Apr 10, 2017

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 :)