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

Oct 17, 2016

$\sin 2 x \cos 2 x$

#### Explanation:

In this exercise we have to apply : two properties
the derivative of product:
$\textcolor{red}{\left(u v\right) ' = u ' \left(x\right) v \left(x\right) + v ' \left(x\right) u \left(x\right)}$

The derivative of a power:
$\textcolor{b l u e}{\left({u}^{n} \left(x\right)\right) ' = n {\left(u\right)}^{n - 1} \left(x\right) u ' \left(x\right)}$

In this exercise let :
$\textcolor{b r o w n}{u \left(x\right) = {\cos}^{2} \left(x\right)}$
$\textcolor{b l u e}{u ' \left(x\right) = 2 \cos x \cos ' x}$
$u ' \left(x\right) = - 2 \cos x \sin x$

Knowing the trigonometric identity that says :
$\textcolor{g r e e n}{\sin 2 x = 2 \sin x \cos x}$

$u ' \left(x\right) = - \textcolor{g r e e n}{\sin 2 x}$

Let:
$\textcolor{b r o w n}{v \left(x\right) = {\sin}^{2} \left(x\right)}$
$\textcolor{b l u e}{v ' \left(x\right) = 2 \sin x \sin ' x}$
$v ' \left(x\right) = 2 \sin x \cos x$
$v ' \left(x\right) = \textcolor{g r e e n}{\sin 2 x}$

So,
$\left({\cos}^{2} x {\sin}^{2} x\right) '$
=color(red)((uv)'
$= \textcolor{red}{u ' \left(x\right) v \left(x\right) + v ' \left(x\right) u \left(x\right)}$
$= \left(- \sin 2 x\right) \left({\sin}^{2} x\right) + \sin \left(2 x\right) {\cos}^{2} x$
$= \sin 2 x \left({\cos}^{2} x - {\sin}^{2} x\right)$
Knowing the trigonometric identity that says :
$\textcolor{g r e e n}{\cos 2 x = {\cos}^{2} x - {\sin}^{2} x}$

Therefore,
$\left({\cos}^{2} x {\sin}^{2} x\right) ' = \sin 2 x \cos 2 x$