# What is the derivative of this function  sin ^2 (2x) + sin (2x+1) ?

Nov 5, 2017

$2 \sin \left(4 x\right) + 2 \cos \left(2 x + 1\right)$

#### Explanation:

${\sin}^{2} \left(2 x\right) = \sin \left(2 x\right) . \sin \left(2 x\right)$

using the product rule

$d \left(u v\right) = u \setminus . \mathrm{dv} + v \setminus . \mathrm{du}$

d(sin(2x).sin(2x)=d(sin(2x)).sin(2x)+d(sin(2x)).sin(2x) ...(a)

$d \left(\sin \left(2 x\right)\right) = 2 \cos \left(2 x\right)$

in a

$2 \cos \left(2 x\right) . \sin \left(2 x\right) + 2 \cos \left(2 x\right) . \sin \left(2 x\right)$

$4 \sin \left(2 x\right) \cos \left(2 x\right)$ it can also writes as $2 \sin \left(4 x\right)$ identity of sin double

for $\sin \left(2 x + 1\right)$

$d \left(\sin \left(u\right)\right) = \cos \left(u\right) . d \left(u\right)$
then

$d \left(\sin \left(2 x + 1\right)\right) = \cos \left(2 x + 1\right) . d \left(2 x + 1\right)$

$d \left(2 x + 1\right) = 2$

$d \left(\sin \left(2 x + 1\right)\right) = 2 \cos \left(2 x + 1\right)$

finally the answer is

$2 \sin \left(4 x\right) + 2 \cos \left(2 x + 1\right)$