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