Answer
$$\frac{{dy}}{{d\theta }} = {\left( {\sin \theta } \right)^{\sqrt \theta }}\left( {\sqrt \theta \cot \theta + \frac{{\ln \left( {\sin \theta } \right)}}{{2\sqrt \theta }}} \right)$$
Work Step by Step
$$\eqalign{
& y = {\left( {\sin \theta } \right)^{\sqrt \theta }} \cr
& {\text{Take the natural logarithm on both sides}} \cr
& \ln y = \ln {\left( {\sin \theta } \right)^{\sqrt \theta }} \cr
& {\text{use power property for logarithms}} \cr
& \ln y = \sqrt \theta \ln \left( {\sin \theta } \right) \cr
& {\text{differentiate both sides with respect to }}\theta \cr
& \frac{1}{y}\frac{{dy}}{{d\theta }} = \sqrt \theta \frac{d}{{d\theta }}\left( {\ln \left( {\sin \theta } \right)} \right) + \ln \left( {\sin \theta } \right)\frac{d}{{d\theta }}\left( {\sqrt \theta } \right) \cr
& \frac{1}{y}\frac{{dy}}{{d\theta }} = \sqrt \theta \left( {\frac{{\cos \theta }}{{\sin \theta }}} \right) + \ln \left( {\sin \theta } \right)\left( {\frac{1}{{2\sqrt \theta }}} \right) \cr
& \frac{1}{y}\frac{{dy}}{{d\theta }} = \sqrt \theta \cot \theta + \frac{{\ln \left( {\sin \theta } \right)}}{{2\sqrt \theta }} \cr
& {\text{solve the equation for }}\frac{{dy}}{{d\theta }} \cr
& \frac{{dy}}{{d\theta }} = y\left( {\sqrt \theta \cot \theta + \frac{{\ln \left( {\sin \theta } \right)}}{{2\sqrt \theta }}} \right) \cr
& {\text{replace }}y = {\left( {\sin \theta } \right)^{\sqrt \theta }} \cr
& \frac{{dy}}{{d\theta }} = {\left( {\sin \theta } \right)^{\sqrt \theta }}\left( {\sqrt \theta \cot \theta + \frac{{\ln \left( {\sin \theta } \right)}}{{2\sqrt \theta }}} \right) \cr} $$
