How do you find the first derivative of ##y=(sinx/(1+cosx))^2##?

Let’s use the , by naming ##u=(sinx)/(1+cosx)##.

The chain rule states that

##(dy)/(dx)=(dy)/(du)(du)/(dx)##

Thus,

##(dy)/(du)=2u##

##(du)/(dx)## – here, we have to apply quocient rule, which states that

Be ##y=(f(x))/(g(x))##, ##(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/(f(x))^2##

##(du)/(dx)=(((cosx)(1+cosx))-(sinx)(-sinx))/(1+cosx)^2##

##(du)/(dx)=((cos^2x+cosx)+sin^2x)/(1+cosx)^2##

##(du)/(dx)=(color(green)(cos^2x+sin^2x)+cosx)/(1+cosx)^2##

##(du)/(dx)=cancel(1+cosx)/(1+cosx)^cancel2##

##(du)/(dx)=1/(1+cosx)##

Now, combining ##(du)/(dx)## and ##(dy)/(du)##:

##(dy)/(dx)=(2u)(1/(1+cosx))##

##(dy)/(dx)=2(sinx/(1+cosx))(1/(1+cosx))##

##(dy)/(dx)=color(green)((2sinx)/(1+cosx)^2)##