Answer
$$\frac{{dy}}{{d\theta }} = \left( {\tan \theta } \right)\sqrt {2\theta + 1} \left( {\frac{{{{\sec }^2}\theta }}{{\tan \theta }} + \frac{1}{{2\theta + 1}}} \right)$$
Work Step by Step
$$\eqalign{
& y = \left( {\tan \theta } \right)\sqrt {2\theta + 1} \cr
& {\text{Take the natural log of both sides:}} \cr
& {\text{Use the properties of logarithms}} \cr
& \ln y = \ln \left( {\left( {\tan \theta } \right)\sqrt {2\theta + 1} } \right) \cr
& {\text{product rule:}} \cr
& \ln y = \ln \left( {\tan \theta } \right) + \ln \left( {\sqrt {2\theta + 1} } \right) \cr
& {\text{power rule:}} \cr
& \ln y = \ln \left( {\tan \theta } \right) + \frac{1}{2}\ln \left( {2\theta + 1} \right) \cr
& {\text{Take derivatives of both sides:}} \cr
& \frac{1}{y}\frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left( {\ln \left( {\tan \theta } \right)} \right) + \frac{1}{2}\frac{d}{{d\theta }}\left( {\ln \left( {2\theta + 1} \right)} \right) \cr
& \frac{1}{y}\frac{{dy}}{{d\theta }} = \frac{{{{\sec }^2}\theta }}{{\tan \theta }} + \frac{1}{2}\left( {\frac{2}{{2\theta + 1}}} \right) \cr
& \frac{1}{y}\frac{{dy}}{{d\theta }} = \frac{{{{\sec }^2}\theta }}{{\tan \theta }} + \frac{1}{2}\left( {\frac{2}{{2\theta + 1}}} \right) \cr
& {\text{solve for }}\frac{{dy}}{{d\theta }} \cr
& \frac{{dy}}{{d\theta }} = y\left( {\frac{{{{\sec }^2}\theta }}{{\tan \theta }} + \frac{1}{{2\theta + 1}}} \right) \cr
& {\text{substitute }}\left( {\tan \theta } \right)\sqrt {2\theta + 1} \sin \theta {\text{ for }}y{\text{ }} \cr
& \frac{{dy}}{{d\theta }} = \left( {\tan \theta } \right)\sqrt {2\theta + 1} \left( {\frac{{{{\sec }^2}\theta }}{{\tan \theta }} + \frac{1}{{2\theta + 1}}} \right) \cr} $$
