Find the derivative of the following function: $\frac{\cos x}{1+\sin x}$

Let $f(x) = \frac{\cos x}{1+\sin x}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:
$f'(x) = \frac{(1+\sin x) \frac{d}{dx}(\cos x) - (\cos x) \frac{d}{dx}(1+\sin x)}{(1+\sin x)^2}$
$f'(x) = \frac{(1+\sin x)(-\sin x) - (\cos x)(\cos x)}{(1+\sin x)^2}$
$f'(x) = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1+\sin x)^2}$
Since $\sin^2 x + \cos^2 x = 1$,we have:
$f'(x) = \frac{-\sin x - 1}{(1+\sin x)^2}$
$f'(x) = \frac{-(1+\sin x)}{(1+\sin x)^2}$
$f'(x) = \frac{-1}{1+\sin x}$