# How do you differentiate sin^2(2x)?

Jul 23, 2016

$2 \sin 4 x$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(A\right)$

here $f \left(g \left(x\right)\right) = {\sin}^{2} \left(2 x\right) = {\left(\sin 2 x\right)}^{2}$

$\Rightarrow f ' \left(g \left(x\right)\right) = 2 \sin 2 x$

Note, 2 applications of $\textcolor{b l u e}{\text{chain rule}}$ required for g'(x)

$g \left(x\right) = \sin 2 x \Rightarrow g ' \left(x\right) = \cos 2 x . \frac{d}{\mathrm{dx}} \left(2 x\right) = 2 \cos 2 x$
$\text{------------------------------------------------------------------}$
Substitute these values into (A)

$2 \sin 2 x .2 \cos 2 x = 4 \sin 2 x \cos 2 x$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sin 4 x = 2 \sin 2 x \cos 2 x} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow \frac{d}{\mathrm{dx}} \left({\sin}^{2} \left(2 x\right)\right) = 4 \sin 2 x \cos 2 x = 2 \sin 4 x$