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How do you find the derivative of # sin ^2 (2x) + sin (2x+1) #?

1 Answer

Konstantinos Michailidis

Mar 9, 2016

It is

#d/dx [sin ^2 (2x) + sin (2x+1)]=2sin2x(d/dx(sin2x))+d/dx[sin(2x+1)] =4*sin2x*cos2x+2*cos(2x+1)#

Finally

#d/dx [sin ^2 (2x) + sin (2x+1)]=4*sin2x*cos2x+2*cos(2x+1)#

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