Answer
$$y = \frac{1}{4}x - \frac{1}{2} + \frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& y = \arctan \frac{x}{2},{\text{ }}\left( {2,\frac{\pi }{4}} \right) \cr
& {\text{Differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\arctan \frac{x}{2}} \right] \cr
& {\text{Use }}\frac{d}{{dx}}\left[ {\arctan u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {x/2} \right)}^2}}}\frac{d}{{dx}}\left[ {\frac{x}{2}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {x/2} \right)}^2}}}\left( {\frac{1}{2}} \right) \cr
& {\text{Calculate the slope at the given point}} \cr
& {\left. {m = \frac{{dy}}{{dx}}} \right|_{x = 2}} = \frac{1}{{1 + {{\left( {2/2} \right)}^2}}}\left( {\frac{1}{2}} \right) \cr
& m = \frac{1}{4} \cr
& {\text{The equation of the tangent line at the point }}\left( {2,\frac{\pi }{4}} \right){\text{ is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - \frac{\pi }{4} = \frac{1}{4}\left( {x - 2} \right) \cr
& y - \frac{\pi }{4} = \frac{1}{4}x - \frac{1}{2} \cr
& y = \frac{1}{4}x - \frac{1}{2} + \frac{\pi }{4} \cr} $$
