Answer
$dy = - \frac{{\sin \theta }}{{2\sqrt {1 + \cos \theta } }}d\theta $
Work Step by Step
$$\eqalign{
& y = \sqrt {1 + \cos \theta } \cr
& {\text{Rewrite the function}} \cr
& y = {\left( {1 + \cos \theta } \right)^{1/2}} \cr
& {\text{Differentiate the function }} \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{{\left( {1 + \cos \theta } \right)}^{1/2}}} \right] \cr
& {\text{Use the chain rule for derivatives}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{2}{\left( {1 + \cos \theta } \right)^{ - 1/2}}\frac{d}{{d\theta }}\left[ {1 + \cos \theta } \right] \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{2}{\left( {1 + \cos \theta } \right)^{ - 1/2}}\left( { - \sin \theta } \right) \cr
& \frac{{dy}}{{d\theta }} = - \frac{{\sin \theta }}{{2{{\left( {1 + \cos \theta } \right)}^{1/2}}}} \cr
& \frac{{dy}}{{d\theta }} = - \frac{{\sin \theta }}{{2\sqrt {1 + \cos \theta } }} \cr
& {\text{Write in the differential form }}dy = f'\left( \theta \right)d\theta \cr
& dy = - \frac{{\sin \theta }}{{2\sqrt {1 + \cos \theta } }}d\theta \cr} $$
