# Question eb80b

Sep 18, 2017

$\frac{x \left(2 - 9 {x}^{3}\right)}{1 - 3 {x}^{3}} ^ \left(\frac{2}{3}\right) .$

#### Explanation:

Suppose that, $y = {x}^{2} {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}} .$

Applying the Product Rule, we get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \cdot \frac{d}{\mathrm{dx}} {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}} + {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2}\right) \ldots . \left(\star\right) .$

Next, we use the Power Rule, and, the Chain Rule, to get,

$\frac{d}{\mathrm{dx}} {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}} = \frac{1}{3} \cdot {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3} - 1} \cdot \frac{d}{\mathrm{dx}} \left(1 - 3 {x}^{3}\right) ,$

$= \frac{1}{3} \cdot {\left(1 - 3 {x}^{3}\right)}^{- \frac{2}{3}} \cdot \left(0 - 3 \cdot 3 {x}^{2}\right) ,$

$\Rightarrow \frac{d}{\mathrm{dx}} {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}} = \frac{- 3 {x}^{2}}{1 - 3 {x}^{3}} ^ \left(\frac{2}{3}\right) \ldots \ldots . \left({\star}_{1}\right) .$

Using $\left({\star}_{1}\right)$ in $\left(\star\right) ,$ we get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \left\{\frac{- 3 {x}^{2}}{1 - 3 {x}^{3}} ^ \left(\frac{2}{3}\right)\right\} + {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}} \cdot 2 x ,$

$= 2 x {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}} - \frac{3 {x}^{4}}{1 - 3 {x}^{3}} ^ \left(\frac{2}{3}\right) ,$

={2x(1-3x^3)^(1/3)(1-3x^3)^(2/3)-3x^4}/((1-3x^3)^(2/3),#

$= \frac{2 x \left(1 - 3 {x}^{3}\right) - 3 {x}^{4}}{1 - 3 {x}^{3}} ^ \left(\frac{2}{3}\right) .$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x \left(2 - 9 {x}^{3}\right)}{1 - 3 {x}^{3}} ^ \left(\frac{2}{3}\right) .$

Enjoy Maths.!

Sep 18, 2017

$\frac{d}{\mathrm{dx}} \left({x}^{2} {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}}\right) = - 3 {x}^{4} {\left(1 - 3 {x}^{3}\right)}^{- \frac{2}{3}} + 2 x {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}}$

#### Explanation:

We have: ${x}^{2} {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}}$

We need to find the derivative of the function $f \left(x\right) = {x}^{2} {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}}$.

Let's begin differentiating $f \left(x\right)$ by applying the product rule:

$R i g h t a r r o w f ' \left(x\right) = {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left({\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}}\right) + {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

Then, let's use the power rule to further differentiate:

$R i g h t a r r o w f ' \left(x\right) = {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left({\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}}\right) + {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}} \cdot 2 x$

Now, the final bit of differentiation must be done by using the chain rule.

Suppose that:

$u = 1 - 3 {x}^{3} R i g h t a r r o w u ' = \frac{d}{\mathrm{dx}} \left(1\right) - \frac{d}{\mathrm{dx}} \left(3 {x}^{3}\right) R i g h t a r r o w u ' = 0 - 3 \cdot 3 {x}^{3 - 1} R i g h t a r r o w u ' = - 9 {x}^{2}$

$\mathmr{and}$

$v = {u}^{\frac{1}{3}} R i g h t a r r o w v ' = \frac{1}{3} {u}^{\frac{1}{3} - 1} R i g h t a r r o w v ' = \frac{1}{3} {u}^{- \frac{2}{3}}$

So:

$R i g h t a r r o w f ' \left(x\right) = {x}^{2} \cdot u ' \cdot v ' + 2 x {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}}$

$R i g h t a r r o w f ' \left(x\right) = {x}^{2} \cdot \left(- 9 {x}^{2}\right) \cdot \frac{1}{3} {u}^{- \frac{2}{3}} + 2 x {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}}$

$R i g h t a r r o w f ' \left(x\right) = - 9 {x}^{4} \cdot \frac{1}{3} {\left(1 - 3 {x}^{3}\right)}^{- \frac{2}{3}} + 2 x {\left(1 - 3 {x}^{3}\right)}^{\frac{1}{3}}$

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