# How do you differentiate the equation?

## ${\left({x}^{3} + 2 x\right)}^{2} = f \left(x\right)$

May 4, 2018

$2 {\left({x}^{3} + 2 x\right)}^{1} \left(3 {x}^{2} + 2\right)$

#### Explanation:

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

Chain Rule:
In general,
$\frac{d}{\mathrm{dx}} {\left[f \left(x\right)\right]}^{n} = n {\left[f \left(x\right)\right]}^{n - 1} \times f ' \left(x\right)$

May 4, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) = 2 \left({x}^{3} + 2 x\right) \left(3 {x}^{2} + 2\right)$

#### Explanation:

Given: $f \left(x\right) = {\left({x}^{3} + 2 x\right)}^{2}$

Set: $y = {\left({x}^{3} + 2 x\right)}^{2}$

Let $u = {x}^{3} + 2 x \implies \frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2} + 2$

Set $y = {u}^{2} \implies \frac{\mathrm{dy}}{\mathrm{du}} = 2 u$

We need $\frac{\mathrm{dy}}{\mathrm{dx}}$ and this is obtained by:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{w h i t e}{\text{dd") dy/(du)color(white)("ddd")xxcolor(white)("dd}} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{w h i t e}{\text{dd") (2u)color(white)("ddd}} \times \left(3 {x}^{2} + 2\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left({x}^{3} + 2 x\right) \times \left(3 {x}^{2} + 2\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) = 2 \left({x}^{3} + 2 x\right) \left(3 {x}^{2} + 2\right)$