Answer
$\frac{{dy}}{{d\theta }} = \cos \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)\left( {1 + {{\sec }^2}\left( {\theta + \cos \theta } \right)\left( {1 - \sin \theta } \right)} \right)$
Work Step by Step
$$\eqalign{
& y = \sin \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {\sin \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)} \right] \cr
& {\text{Use }}\frac{d}{{d\theta }}\left[ {\sin u} \right] = \cos u\frac{{du}}{{d\theta }} \cr
& \frac{{dy}}{{d\theta }} = \cos \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)\frac{d}{{d\theta }}\left[ {\theta + \tan \left( {\theta + \cos \theta } \right)} \right] \cr
& \frac{{dy}}{{d\theta }} = \cos \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)\left( {\frac{d}{{d\theta }}\left[ \theta \right] + \frac{d}{{d\theta }}\left[ {\tan \left( {\theta + \cos \theta } \right)} \right]} \right) \cr
& {\text{Use }}\frac{d}{{d\theta }}\left[ {\tan u} \right] = {\sec ^2}u\frac{{du}}{{d\theta }} \cr
& \frac{{dy}}{{d\theta }} = \cos \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)\left( {1 + {{\sec }^2}\left( {\theta + \cos \theta } \right)\left[ {\theta + \cos \theta } \right]'} \right) \cr
& {\text{Compute derivatives}} \cr
& \frac{{dy}}{{d\theta }} = \cos \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)\left( {1 + {{\sec }^2}\left( {\theta + \cos \theta } \right)\left( {1 - \sin \theta } \right)} \right) \cr} $$
