Answer
$$\frac{{dy}}{{dx}} = \frac{1}{{x\left( {1 + {{\ln }^2}x} \right)}}$$
Work Step by Step
$$\eqalign{
& y = {\tan ^{ - 1}}\left( {\ln x} \right) \cr
& {\text{find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{{d\left( {{{\tan }^{ - 1}}\left( {\ln x} \right)} \right)}}{{dx}} \cr
& {\text{ we can use the formula }}\cr
& \frac{{d\left( {{{\tan }^{ - 1}}u} \right)}}{{dx}} = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}}.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr
& {\text{here }}u = \ln x;{\text{then}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {\ln x} \right)}^2}}}\frac{{d\left( {\ln x} \right)}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {\ln x} \right)}^2}}}\left( {\frac{1}{x}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{x\left( {1 + {{\ln }^2}x} \right)}} \cr} $$
