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