I am not sure how you would approach this using the chain rule, but I can use the quotient rule.
The derivative of cos(u)=-sin(u)
The derivative of sin(u)=cos(u)
The quotient rule states:
[f'(x)g(x)-f(x)g'(x)] / [g(x)]^2
f'(x)= -sin(x)
g(x)= [1+sin(x)]
f(x)= cos(x)
g'(x)= cos(x)
Plug in:
[-sin(x)*[1+sin(x)] - [cos(x)*cos(x)] / (1+sin(x))^2
Simplify by distributing
-sin(x) - sin^2(x) - [cos(x)*cos(x)] / (1+sin(x))^2
Multiply the cos(x)*cos(x)
-sin(x) - sin^2(x) - cos^2(x) / (1+sin(x))^2
That is the answer or the answer could be simplified further by pulling out the negative.
- [sin(x)+sin^2(x)+cos^2(x)] / (1+sin(x))^2
