Answer
$$\frac{{dy}}{{d\theta }} = 2\cos \left( {\ln \theta } \right)$$
Work Step by Step
$$\eqalign{
& y = \theta \left( {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {\theta \left( {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right)} \right] \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{d\theta }} = \theta \frac{d}{{d\theta }}\left[ {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right] + \left( {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right)\frac{d}{{d\theta }}\left[ \theta \right] \cr
& {\text{use the rules }}\cr
& \frac{d}{{d\theta }}\left[ {\sin u} \right] = \cos u\frac{{du}}{{d\theta }}{\text{ and }}\frac{d}{{d\theta }}\left[ {\cos \theta } \right] = - \sin \theta \frac{{du}}{{d\theta }} \cr
& \frac{{dy}}{{d\theta }} = \theta \left( {\cos \left( {\ln \theta } \right)\frac{d}{{d\theta }}\left[ {\ln \theta } \right] - \sin \left( {\ln \theta } \right)\frac{d}{{d\theta }}\left[ {\ln \theta } \right]} \right) + \left( {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right)\frac{d}{{d\theta }}\left[ \theta \right] \cr
& {\text{solve the derivatives}} \cr
& \frac{{dy}}{{d\theta }} = \theta \left( {\cos \left( {\ln \theta } \right)\left( {\frac{1}{\theta }} \right) - \sin \left( {\ln \theta } \right)\left( {\frac{1}{\theta }} \right)} \right) + \left( {\sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right)} \right)\left( 1 \right) \cr
& {\text{simplifying, we get:}} \cr
& \frac{{dy}}{{d\theta }} = \cos \left( {\ln \theta } \right) - \sin \left( {\ln \theta } \right) + \sin \left( {\ln \theta } \right) + \cos \left( {\ln \theta } \right) \cr
& \frac{{dy}}{{d\theta }} = 2\cos \left( {\ln \theta } \right) \cr} $$
