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

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Answer 1 · 2017-11-05T17:52:25

$2 sin (4 x) + 2 cos (2 x + 1)$ $2sin(4x)+2cos(2x+1)$

${sin}^{2} (2 x) = sin (2 x) . sin (2 x)$ $sin^2(2x)=sin(2x).sin(2x)$

$d (u v) = u . d v + v . d u$ $d(uv)=u.dv+v.du$

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

$d (sin (2 x)) = 2 cos (2 x)$ $d(sin(2x))=2cos(2x)$

in a

$2 cos (2 x) . sin (2 x) + 2 cos (2 x) . sin (2 x)$ $2cos(2x).sin(2x)+2cos(2x).sin(2x)$

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

for $sin (2 x + 1)$ $sin(2x+1)$

$d (sin (u)) = cos (u) . d (u)$ $d(sin(u))=cos(u).d(u)$
then

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

$d (2 x + 1) = 2$ $d(2x+1)=2$

$d (sin (2 x + 1)) = 2 cos (2 x + 1)$ $d(sin(2x+1))=2cos(2x+1)$

finally the answer is

$2 sin (4 x) + 2 cos (2 x + 1)$ $2sin(4x)+2cos(2x+1)$

What is the derivative of this function sin ^2 (2x) + sin (2x+1) sin2(2x)+sin(2x+1) sin ^2 (2x) + sin (2x+1) ?