Answer
The Derivative is:
$y'=-\frac{1}{(1+\sin x)}$
Work Step by Step
$y=\frac{\cos x}{1+\sin x}$
Applying Derivative Rules:
$y'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^2(x)}$
$y'=\frac{\frac{d}{dx}(\cos x)\cdot (1+\sin x) - (\cos x)\cdot \frac{d}{dx}(1+\sin x)}{(1+\sin x)^2}$
$y'=\frac{(-\sin x)(1+\sin x)-(\cos x)(\cos x)}{(1+\sin x)^2}$
$y'=\frac{-\sin x-\sin^2 x -\cos^2 x}{(1+\sin x)^2}$
$y'=\frac{(-1)(\sin x+\sin^2 x+\cos^2 x)}{(1+\sin x)^2}$
Applying trigonometric identities
$\sin^2 x+\cos^2 x=1$
$y'=\frac{-(1 +\sin x)}{(1+\sin x)^2}$
$y'=-\frac{1}{(1+\sin x)}$
