Question

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

Answer 1

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)`