# How do you find the derivative of y= ( 2x sin x ) ^ 3x cos x?

Dec 8, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 x {\left(2 x \sin x\right)}^{2} \sin x \cos x + {\left(2 x \sin x\right)}^{3} \cos x - x {\left(2 x \sin x\right)}^{3} \sin x$

#### Explanation:

$\left[f g h\right] ' = f ' g h + f g ' h + f g h '$

In this case, let $f = {\left(2 x \sin x\right)}^{3}$, $g = x$, and $h = \cos x$
$y ' = \textcolor{red}{\left[{\left(2 x \sin x\right)}^{3}\right] '} x \cos x + {\left(2 x \sin x\right)}^{3} \textcolor{g r a y}{x '} \cos x + {\left(2 x \sin x\right)}^{3} x \textcolor{p u r p \le}{\left[\cos x\right] '}$

To find $\textcolor{red}{\left[{\left(2 x \sin x\right)}^{3}\right] '}$ we need to use the chain rule
$\left[f \left(g\right)\right] ' = f ' \left(g\right) \times g '$

In this case, let $f = {g}^{3}$ and $g = 2 x \sin x$
$\textcolor{red}{\left[f \left(g\right)\right] ' = 3 {\left(2 x \sin x\right)}^{2} \times} \textcolor{g r e e n}{\left[2 x \sin x\right] '}$

To find $\textcolor{g r e e n}{\left[2 x \sin x\right] '}$ we need to use the product rule for 2 factors
$\left[f g\right] ' = f ' g + f g '$

In this case, let $f = 2 x$ and $g = \sin x$
$\textcolor{g r e e n}{\left[f g\right] ' = 2 \sin x + 2 x \cos x}$

$\textcolor{red}{\left[f \left(g\right)\right] ' = 3 {\left(2 x \sin x\right)}^{2} \times} \textcolor{g r e e n}{2 \sin x + 2 x \cos x}$
$\textcolor{red}{\left[{\left(2 x \sin x\right)}^{3}\right] ' = 3 {\left(2 x \sin x\right)}^{2} \times 2 \sin x + 2 x \cos x}$

$\textcolor{g r a y}{x '}$ is easy to find
$\textcolor{g r a y}{x ' = 1}$

$\textcolor{p u r p \le}{\left[\cos x\right] '}$ is also something you should know
color(purple)([cosx]'=-sinx

$y ' = \textcolor{red}{\left[3 {\left(2 x \sin x\right)}^{2} \times 2 \sin x + 2 x \cos x\right]} x \cos x + {\left(2 x \sin x\right)}^{3} \textcolor{g r a y}{\times 1 \times} \cos x + {\left(2 x \sin x\right)}^{3} x \textcolor{p u r p \le}{\left(- \sin x\right)}$

Depending on your teacher, that may be as far as you need to go but some will want it simplified, which I will do below.

$y ' = 6 x {\left(2 x \sin x\right)}^{2} \sin x \cos x + {\left(2 x \sin x\right)}^{3} \cos x - x {\left(2 x \sin x\right)}^{3} \sin x$

Equivalent forms:

$y ' = {\left(2 x \sin x\right)}^{2} \left(6 x \sin x \cos x + 2 x \sin x \cos x - 2 {x}^{2} {\sin}^{2} x\right)$
$y ' = {\left(2 x \sin x\right)}^{3} \left(3 \cos x + \cos x - x \sin x\right)$
$y ' = {\left(2 x \sin x\right)}^{3} \left(4 \cos x - x \sin x\right)$