Answer
$$\frac{{dy}}{{dx}} = \frac{2}{{\pi \left( {1 + 4{x^2}} \right)}}$$
Work Step by Step
\lim\limits_{a \to b}$$\eqalign{
& y = \frac{1}{\pi }{\tan ^{ - 1}}2x \cr
& {\text{find the derivative}} \cr
& y' = \left( {\frac{1}{\pi }{{\tan }^{ - 1}}2x} \right)' \cr
& y' = \frac{1}{\pi }\left( {{{\tan }^{ - 1}}2x} \right)' \cr
& {\text{differentiate using the formula }} \cr
& \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}}{\text{ }}\left( {{\text{see page 467}}} \right) \cr
& \cr
& \frac{{dy}}{{dx}} = \frac{1}{\pi }\left( {\frac{1}{{1 + {{\left( {2x} \right)}^2}}}} \right)\left( {2x} \right)' \cr
& \frac{{dy}}{{dx}} = \frac{1}{\pi }\left( {\frac{1}{{1 + 4{x^2}}}} \right)\left( 2 \right) \cr
& {\text{simplifying}}{\text{, }} \cr
& \frac{{dy}}{{dx}} = \frac{2}{{\pi \left( {1 + 4{x^2}} \right)}} \cr} $$
