Answer
$$\,\,\,\,\,y = x - \frac{1}{2} + \frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\tan ^{ - 1}}2x;{\text{ }}\left( {\frac{1}{2},\frac{\pi }{4}} \right) \cr
& {\text{differentiate }}f\left( x \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}2x} \right] \cr
& {\text{use }}\frac{d}{{du}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}} \cr
& f'\left( x \right) = \frac{1}{{1 + {{\left( {2x} \right)}^2}}}\frac{d}{{dx}}\left[ {2x} \right] \cr
& f'\left( x \right) = \frac{1}{{1 + 4{x^2}}}\left( 2 \right) \cr
& f'\left( x \right) = \frac{2}{{1 + 4{x^2}}} \cr
& \cr
& {\text{find the slope at the point }}\left( {\frac{1}{2},\frac{\pi }{4}} \right) \cr
& \,\,m = f'\left( {\frac{1}{2}} \right) = \frac{2}{{1 + 4{{\left( {1/2} \right)}^2}}} \cr
& \,\,m = 1 \cr
& \,{\text{find the equation of the tangent line at the point }}\left( {\frac{1}{2},\frac{\pi }{4}} \right) \cr
& \,\,\,\,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \,\,\,\,\,y - \frac{\pi }{4} = \left( {x - \frac{1}{2}} \right) \cr
& \,\,\,\,\,y - \frac{\pi }{4} = x - \frac{1}{2} \cr
& \,\,\,\,\,y = x - \frac{1}{2} + \frac{\pi }{4} \cr} $$
