Question

Answer 1

1. 

a)y'=2sinx.cosx-3sin3x=sin2x-3sin3x

b)y'=\(\dfrac{\left(cosx\right)'\left(2+sin2x\right)-cosx\left(2+sin2x\right)'}{\left(2+sin2x\right)^2}\)=\(\dfrac{-2sinx-sinxsin2x-2cosxcos2x}{\left(2+sin2x\right)^2}\)

Answer 2

2. y'=-2sin2x.sin3x+cos2x.3cos3x

y'=\(\dfrac{-2sinx}{2\sqrt{3+2cosx}}\)