Answer
$$\frac{{dy}}{{d\theta }} = \frac{1}{{2\theta \left( {1 + \sqrt \theta } \right)}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\frac{{\sqrt \theta }}{{1 + \sqrt \theta }}} \right) \cr
& {\text{using the logarithmic property ln}}\left( {\frac{a}{b}} \right) = \ln a - \ln b \cr
& y = \ln \left( {\sqrt \theta } \right) - \ln \left( {1 + \sqrt \theta } \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {\ln \left( {\sqrt \theta } \right)} \right] - \frac{d}{{d\theta }}\left[ {\ln \left( {1 + \sqrt \theta } \right)} \right] \cr
& {\text{use the rule }}\frac{d}{{d\theta }}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{d\theta }}.{\text{ then}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\sqrt \theta }}\frac{d}{{d\theta }}\left[ {\sqrt \theta } \right] - \frac{1}{{1 + \sqrt \theta }}\frac{d}{{d\theta }}\left[ {1 + \sqrt \theta } \right] \cr
& {\text{solving derivatives}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\sqrt \theta }}\left( {\frac{1}{{2\sqrt \theta }}} \right) - \frac{1}{{1 + \sqrt \theta }}\left( {\frac{1}{{2\sqrt \theta }}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{2\sqrt \theta \sqrt \theta }} - \frac{1}{{2\sqrt \theta \left( {1 + \sqrt \theta } \right)}} \cr
& {\text{factoring}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{2\sqrt \theta }}\left( {\frac{1}{{\sqrt \theta }} - \frac{1}{{1 + \sqrt \theta }}} \right) \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{2\sqrt \theta }}\left( {\frac{{1 + \sqrt \theta - \sqrt \theta }}{{\sqrt \theta \left( {1 + \sqrt \theta } \right)}}} \right) \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{2\theta }}\left( {\frac{1}{{1 + \sqrt \theta }}} \right) \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{2\theta \left( {1 + \sqrt \theta } \right)}} \cr} $$
