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