a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$
b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$
c. $y'=3sin^2(2x+1)[sin(2x+1)]'=3sin^2(2x+1)[cos(2x+1).(2x+1)']=6sin^2(2x+1).cos(2x+1)=3sin(4x+2).sin(2x+1)$