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

1 Answer
May 22, 2015

Let's use the chain rule, 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)#