Answer
$$\frac{{dy}}{{dx}} = 0$$
Work Step by Step
$$\eqalign{
& y = {\cot ^{ - 1}}\frac{1}{x} - {\tan ^{ - 1}}x \cr
& y = {\cot ^{ - 1}}{x^{ - 1}} - {\tan ^{ - 1}}x \cr
& {\text{find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{{d\left( {{{\cot }^{ - 1}}{x^{ - 1}}} \right)}}{{dx}} - \frac{{d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}} \cr
& {\text{ use the formulas }} \cr
& \frac{{d\left( {{{\tan }^{ - 1}}u} \right)}}{{dx}} = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}}\cr
& \frac{{d\left( {{{\cot }^{ - 1}}u} \right)}}{{dx}} = - \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}}.\,\,\left( {{\text{see table 7}}{\text{.3}}} \right) \cr
& {\text{then}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{1 + {{\left( {{x^{ - 1}}} \right)}^2}}}\frac{{d\left( {{x^{ - 1}}} \right)}}{{dx}} - \frac{1}{{1 + {x^2}}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{1 + {x^{ - 2}}}}\left( { - \frac{1}{{{x^2}}}} \right) - \frac{1}{{1 + {x^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\frac{{{x^2} + 1}}{{{x^2}}}}}\left( { - \frac{1}{{{x^2}}}} \right) - \frac{1}{{1 + {x^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{{x^2} + 1}} - \frac{1}{{1 + {x^2}}} \cr
& \frac{{dy}}{{dx}} = 0 \cr} $$
