Answer
$$dy = \frac{{2 - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx$$
Work Step by Step
$$\eqalign{
& y = \frac{{2x}}{{1 + {x^2}}} \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{2x}}{{1 + {x^2}}}} \right] \cr
& {\text{Use the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 + {x^2}} \right)\left( 2 \right) - 2x\left( {2x} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{2 + 2{x^2} - 4{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{{2 - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr
& {\text{Write in differential form}} \cr
& dy = \frac{{2 - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx \cr} $$
