Answer
$$\frac{1}{{\theta + 1}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {2\theta + 2} \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {\ln \left( {2\theta + 2} \right)} \right] \cr
& {\text{use the formula }}\frac{d}{{d\theta }}\ln u = \frac{1}{u}\frac{{du}}{{d\theta }}{\text{, }}u > 0\cr
& {\text{ where }}u{\text{ is any differentiable function of }}\theta \cr
& {\text{for this exercise let }}u = \theta + 1{\text{; then}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{2\theta + 2}}\frac{d}{{d\theta }}\left[ {2\theta + 2} \right] \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{2\theta + 2}}\left( 2 \right) \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{d\theta }} = \frac{2}{{2\left( {\theta + 1} \right)}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\theta + 1}} \cr} $$
