# Question #91477

Jan 7, 2018

$3 {\sin}^{2} \left(x\right) \cos \left(x\right)$

#### Explanation:

${\sin}^{3} \left(x\right) \iff {\left(\sin x\right)}^{3}$

Using the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = \sin \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({\sin}^{3} \left(x\right)\right) = \frac{\mathrm{dy}}{\mathrm{du}} {\left(u\right)}^{3} \cdot \frac{\mathrm{du}}{\mathrm{dx}} \left(u\right)$

$\textcolor{w h i t e}{88888888888} = 3 {u}^{2} \cdot \cos \left(x\right)$

$u = \sin \left(x\right)$

$\textcolor{w h i t e}{88888888888} = 3 {\sin}^{2} \left(x\right) \cos \left(x\right)$

Jan 7, 2018

$3 {\sin}^{2} \left(x\right) \cos \left(x\right)$

#### Explanation:

Using the chain rule,

$\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = \sin \left(x\right)$

$\frac{\mathrm{du}}{\mathrm{dx}} = \cos \left(x\right)$

$f = {u}^{3}$

$\frac{\mathrm{df}}{\mathrm{du}} = 3 {u}^{2}$

$\therefore \frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = 3 {u}^{2} \cos \left(x\right)$

Substitute $u = \sin \left(x\right)$ back, we get

$\therefore \frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = 3 {\sin}^{2} \left(x\right) \cos \left(x\right)$

Jan 7, 2018

$3 {\sin}^{2} x \cos x$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$\text{express } {\sin}^{3} x = {\left(\sin x\right)}^{3}$

$\Rightarrow \frac{d}{\mathrm{dx}} \left({\left(\sin x\right)}^{3}\right)$

$= 3 {\left(\sin x\right)}^{2} \times \frac{d}{\mathrm{dx}} \left(\sin x\right) = 3 {\sin}^{2} x \cos x$