Derivative using chain rule of cosx/1+sinx

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