What is the derivative of sin(2x)cos(2x)?

$2 \setminus \cos \left(4 x\right)$

Explanation:

Given function:

$\setminus \sin \left(2 x\right) \setminus \cos \left(2 x\right)$

$\frac{1}{2} \left(2 \setminus \sin \left(2 x\right) \setminus \cos \left(2 x\right)\right)$

$\frac{1}{2} \setminus \sin \left(4 x\right)$

Differentiating given function w.r.t. $x$ as follows

$\frac{d}{\mathrm{dx}} \left(\frac{1}{2} \setminus \sin \left(4 x\right)\right)$

$= \frac{1}{2} \setminus \frac{d}{\mathrm{dx}} \left(\setminus \sin \left(4 x\right)\right)$

$= \frac{1}{2} \setminus \cos \left(4 x\right) \frac{d}{\mathrm{dx}} \left(4 x\right)$

$= \frac{1}{2} \setminus \cos \left(4 x\right) \left(4\right)$

$= 2 \setminus \cos \left(4 x\right)$

Jul 21, 2018

$2 \cos 4 x$

Explanation:

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

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

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

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

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

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

$= - 2 {\sin}^{2} 2 x + 2 {\cos}^{2} 2 x$

$= 2 \left({\cos}^{2} 2 x - {\sin}^{2} 2 x\right) = 2 \cos 4 x$