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

Ifound this on google “that means that the graph of a function with direct variation has these features: it is a straight line: because the slope is the ratio y/x = k. it passes through the origin (0,0): because y = kx = k (0) = 0.” i hope this you for what your looking for