## Thomas' Calculus 13th Edition

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