Answer
$\frac{{dy}}{{dx}} = \frac{{ - \sin x - \cos \left( {x + y} \right)}}{{\cos \left( {x + y} \right) + \sin y}}$
Work Step by Step
$$\eqalign{
& \sin \left( {x + y} \right) = \cos x + \cos y \cr
& {\text{Differentiating both sides of the equation with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\sin \left( {x + y} \right)} \right] = \frac{d}{{dx}}\left[ {\cos x} \right] + \frac{d}{{dx}}\left[ {\cos y} \right] \cr
& y{\text{ is a function of }}x,{\text{ using the chain rule we have}} \cr
& \cos \left( {x + y} \right)\frac{d}{{dx}}\left[ {x + y} \right] = - \sin x - \sin y\frac{{dy}}{{dx}} \cr
& \cos \left( {x + y} \right)\left( {1 + \frac{{dy}}{{dx}}} \right) = - \sin x - \sin y\frac{{dy}}{{dx}} \cr
& \cos \left( {x + y} \right) + \cos \left( {x + y} \right)\frac{{dy}}{{dx}} = - \sin x - \sin y\frac{{dy}}{{dx}} \cr
& {\text{Collect the terms that contains }}\frac{{dy}}{{dx}} \cr
& \cos \left( {x + y} \right)\frac{{dy}}{{dx}} + \sin y\frac{{dy}}{{dx}} = - \sin x - \cos \left( {x + y} \right) \cr
& {\text{Solve this equation for }}\frac{{dy}}{{dx}} \cr
& \left( {\cos \left( {x + y} \right) + \sin y} \right)\frac{{dy}}{{dx}} = - \sin x - \cos \left( {x + y} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{ - \sin x - \cos \left( {x + y} \right)}}{{\cos \left( {x + y} \right) + \sin y}} \cr} $$
