How do you differentiate f(x)= x^3 (1-3x^2)^4  using the product rule?

Aug 12, 2018

$f ' \left(x\right) = 3 {x}^{2} {\left(1 - 3 {x}^{2}\right)}^{3} \left(1 - 11 {x}^{2}\right)$

Explanation:

Here ,

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

Diff.w.r.t $x$ ,using product rule:

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

Diff.w.r.t. $x$ using chain rule:

$f ' \left(x\right) = {x}^{3} 4 {\left(1 - 3 {x}^{2}\right)}^{3} \left(- 6 x\right) + {\left(1 - 3 {x}^{2}\right)}^{4} \left(3 {x}^{2}\right)$

$\therefore f ' \left(x\right) = {\left(1 - 3 {x}^{2}\right)}^{3} \left\{- 24 {x}^{4} + \left(1 - 3 {x}^{2}\right) 3 {x}^{2}\right\}$

$\therefore f ' \left(x\right) = {\left(1 - 3 {x}^{2}\right)}^{3} \left\{- 24 {x}^{4} + 3 {x}^{2} - 9 {x}^{4}\right\}$

$\therefore f ' \left(x\right) = {\left(1 - 3 {x}^{2}\right)}^{3} \left(3 {x}^{2} - 33 {x}^{4}\right)$

$\therefore f ' \left(x\right) = 3 {x}^{2} {\left(1 - 3 {x}^{2}\right)}^{3} \left(1 - 11 {x}^{2}\right)$