Answer
$$\frac{{dy}}{{dx}} = \frac{{2x}}{{2 + {x^2}}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {2 + {x^2}} \right) \cr
& {\text{differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln \left( {2 + {x^2}} \right)} \right] \cr
& {\text{use }}\frac{d}{{dx}}\ln g\left( x \right) = \frac{1}{{g\left( x \right)}}g'\left( x \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2 + {x^2}}}\frac{d}{{dx}}\left[ {2 + {x^2}} \right] \cr
& {\text{find derivative}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2 + {x^2}}}\left( {2x} \right) \cr
& {\text{multiply}} \cr
& \frac{{dy}}{{dx}} = \frac{{2x}}{{2 + {x^2}}} \cr} $$
