Answer
$$\frac{{dy}}{{dx}} = - \frac{1}{{\left( {1 + {x^2}} \right){{\left( {{{\tan }^{ - 1}}x} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{{{\tan }^{ - 1}}x}} \cr
& {\text{write with negative exponent }}\frac{1}{{{u^n}}} = {u^{ - n}} \cr
& y = {\left( {{{\tan }^{ - 1}}x} \right)^{ - 1}} \cr
& {\text{differentiate using the chain rule}} \cr
& \frac{{dy}}{{dx}} = - {\left( {{{\tan }^{ - 1}}x} \right)^{ - 2}}\left( {{{\tan }^{ - 1}}x} \right)' \cr
& {\text{differentiate using the formula }} \cr
& \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}}{\text{ }}\left( {{\text{see page 467}}} \right) \cr
& \frac{{dy}}{{dx}} = - {\left( {{{\tan }^{ - 1}}x} \right)^{ - 2}}\left( {\frac{1}{{1 + {x^2}}}} \right) \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\left( {1 + {x^2}} \right){{\left( {{{\tan }^{ - 1}}x} \right)}^2}}} \cr} $$
