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