Answer
$$\frac{{dy}}{{d\theta }} = \frac{{\theta + 5}}{{\theta \cos \theta }}\left( {\frac{1}{{\theta + 5}} - \frac{1}{\theta } + \frac{{\sin \theta }}{{\cos \theta }}} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{{\theta + 5}}{{\theta \cos \theta }} \cr
& {\text{Take the natural logarithm of both sides:}} \cr
& {\text{Use the properties of logarithms}} \cr
& \ln y = \ln \left( {\frac{{\theta + 5}}{{\theta \cos \theta }}} \right) \cr
& {\text{quotient rule:}} \cr
& \ln y = \ln \left( {\theta + 5} \right) - \ln \left( {\theta \cos \theta } \right) \cr
& {\text{product rule:}} \cr
& \ln y = \ln \left( {\theta + 5} \right) - \ln \left( \theta \right) - \ln \left( {\cos \theta } \right) \cr
& {\text{Take derivatives of both sides with respect to }}\theta \cr
& \frac{1}{y}\frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left( {\ln \left( {\theta + 5} \right)} \right) - \frac{d}{{d\theta }}\left( {\ln \left( \theta \right)} \right) - \frac{d}{{d\theta }}\left( {\ln \left( {\cos \theta } \right)} \right) \cr
& \frac{1}{y}\frac{{dy}}{{d\theta }} = \frac{1}{{\theta + 5}} - \frac{1}{\theta } - \frac{{ - \sin \theta }}{{\cos \theta }} \cr
& \frac{1}{y}\frac{{dy}}{{d\theta }} = \frac{1}{{\theta + 5}} - \frac{1}{\theta } + \frac{{\sin \theta }}{{\cos \theta }} \cr
& {\text{solve for }}\frac{{dy}}{{d\theta }} \cr
& \frac{{dy}}{{d\theta }} = y\left( {\frac{1}{{\theta + 5}} - \frac{1}{\theta } + \frac{{\sin \theta }}{{\cos \theta }}} \right) \cr
& {\text{substitute }}\frac{{\theta + 5}}{{\theta \cos \theta }}\sin \theta {\text{ for }}y{\text{ }} \cr
& \frac{{dy}}{{d\theta }} = \frac{{\theta + 5}}{{\theta \cos \theta }}\left( {\frac{1}{{\theta + 5}} - \frac{1}{\theta } + \frac{{\sin \theta }}{{\cos \theta }}} \right) \cr} $$
