## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{d\theta }} = \sqrt {\theta + 3} \sin \theta \left( {\frac{1}{{2\left( {\theta + 3} \right)}} + \cot \theta } \right)$$
\eqalign{ & y = \sqrt {\theta + 3} \sin \theta \cr & {\text{Take the natural logarithm of both sides:}} \cr & {\text{Use the properties of logarithms}} \cr & \ln y = \ln \left( {\sqrt {\theta + 3} \sin \theta } \right) \cr & {\text{product rule:}} \cr & \ln y = \ln \left( {\sqrt {\theta + 3} } \right) + \ln \left( {\sin \theta } \right) \cr & {\text{power rule:}} \cr & \ln y = \frac{1}{2}\ln \left( {\theta + 3} \right) + \ln \left( {\sin \theta } \right) \cr & {\text{Take derivatives of both sides with respect to }}\theta \cr & \frac{1}{y}\frac{{dy}}{{d\theta }} = \frac{1}{2}\frac{d}{{d\theta }}\left( {\ln \left( {\theta + 3} \right)} \right) + \frac{d}{{d\theta }}\left( {\ln \left( {\sin \theta } \right)} \right) \cr & \frac{1}{y}\frac{{dy}}{{d\theta }} = \frac{1}{2}\left( {\frac{1}{{\theta + 3}}} \right) + \frac{{\cos \theta }}{{\sin \theta }} \cr & {\text{solve for }}\frac{{dy}}{{d\theta }} \cr & \frac{{dy}}{{d\theta }} = y\left( {\frac{1}{{2\left( {\theta + 3} \right)}} + \frac{{\cos \theta }}{{\sin \theta }}} \right) \cr & {\text{substitute }}\sqrt {\theta + 3} \sin \theta {\text{ for }}y{\text{ }} \cr & \frac{{dy}}{{d\theta }} = \sqrt {\theta + 3} \sin \theta \left( {\frac{1}{{2\left( {\theta + 3} \right)}} + \cot \theta } \right) \cr}