How do you differentiate #sin ^2 (2x) + sin (2x+1) #?

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Answer 1 · 2017-12-10T14:32:48

The answer is #=sin(4x)+2cos(2x+1)#

We need

#sin(u(x))'=u'(x)cos(u(x))#

#sin2a=2sinacosa#

Let #y=sin^2(2x)+sin(2x+1)#

Then, differentiating with respect to #x# (applying the chain rule)

#dy/dx=2sin(2x)*cos(2x)+2cos(2x+1)#

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