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)