#### Answer

$\displaystyle y' = \frac{-1}{(1+sinx)}$

#### Work Step by Step

$\displaystyle y = \frac{cosx}{1+sinx}$
$\displaystyle y' = \frac{-(1+sinx)sinx - cos^{2}x}{(1+sinx)^{2}}$
$\displaystyle y' = \frac{-sinx -sin^{2}x- cos^{2}x}{(1+sinx)^{2}}$
$\displaystyle y' = \frac{-(1+sinx)}{(1+sinx)^{2}}$
$\displaystyle y' = \frac{-1}{(1+sinx)}$