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

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Answer 1 · 2017-11-05T17:52:25

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

#sin^2(2x)=sin(2x).sin(2x)#

#d(uv)=u\.dv+v\.du#

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

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

in a

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

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

for #sin(2x+1)#

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

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

#d(2x+1)=2#

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

finally the answer is

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