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. $y'=3sin^2(2x+1)[sin(2x+1)]'=2sin^2(2x+1)[cos(2x+1).(2x+1)']=4sin^2(2x+1).cos(2x+1)=2sin(4x+2).sin(2x+1)$
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. $y'=2sin(2x+1)[sin(2x+1)]=2sin(2x+1)[cos(2x+1).(2x+1)']=4sin(2x+1).cos(2x+1)=2sin(4x+2)$
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. $y'=
3sin
^2(2x+1)[sin(2x+1)]
'=2sin
^2(2x+1)[cos(2x+1).(2x+1)']=4sin
^2(2x+1).cos(2x+1)=2sin(4x+2
).sin(2x+1)$