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)

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

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

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