# Question #6ae3e

Apr 10, 2017

$f ' \left(x\right) = 4 {\left(\frac{{x}^{3} - 1}{2 {x}^{3} + 1}\right)}^{3} \times \frac{9 {x}^{2}}{{\left(2 {x}^{3} - 1\right)}^{2}}$

#### Explanation:

Assuming that your looking for the derivative of;
$f \left(x\right) = {\left[\frac{{x}^{3} - 1}{2 {x}^{3} + 1}\right]}^{4}$
Let $u \left(x\right) = \frac{{x}^{3} - 1}{2 {x}^{3} + 1}$

$\therefore$ $f = {u}^{4}$

and

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ (chain rule)

Using quotient rule to differentiate $u \left(x\right)$
REMEMBER - Quotient Rule
$f \left(x\right) = \frac{u}{v}$
$f ' \left(x\right) = \frac{u ' v - u v '}{v} ^ 2$

$u ' \left(x\right) = \frac{\left(3 {x}^{2}\right) \left(2 {x}^{3} + 1\right) - \left(6 {x}^{2}\right) \left({x}^{3} - 1\right)}{{\left(2 {x}^{3} - 1\right)}^{2}}$
$u ' \left(x\right) = \frac{\left(6 {x}^{5} + 3 {x}^{2}\right) - \left(6 {x}^{5} - 6 {x}^{2}\right)}{{\left(2 {x}^{3} - 1\right)}^{2}}$
$u ' \left(x\right) = \frac{9 {x}^{2}}{{\left(2 {x}^{3} - 1\right)}^{2}}$

and, as $f = {u}^{4}$, $f ' = 4 {u}^{3}$

$\therefore$ $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = \left(4 {u}^{3}\right) \frac{9 {x}^{2}}{{\left(2 {x}^{3} - 1\right)}^{2}}$
Substituting back in $u \left(x\right) = \frac{{x}^{3} - 1}{2 {x}^{3} + 1}$
$f ' \left(x\right) = 4 {\left(\frac{{x}^{3} - 1}{2 {x}^{3} + 1}\right)}^{3} \times \frac{9 {x}^{2}}{{\left(2 {x}^{3} - 1\right)}^{2}}$

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

hope this helps :)