Answer
$$\frac{{dy}}{{dx}} = - \frac{1}{{x\sqrt {{x^2} - 1} }}$$
Work Step by Step
$$\eqalign{
& y = {\sin ^{ - 1}}\left( {\frac{1}{x}} \right) \cr
& {\text{differentiate using the formula }} \cr
& \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}}{\text{ }}\left( {{\text{see page 467}}} \right) \cr
& \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {{\left( {\frac{1}{x}} \right)}^2}} }}\frac{d}{{dx}}\left( {\frac{1}{x}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - \frac{1}{{{x^2}}}} }}\left( { - \frac{1}{{{x^2}}}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {\frac{{{x^2} - 1}}{{{x^2}}}} }}\left( { - \frac{1}{{{x^2}}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left( {1/x} \right)\sqrt {{x^2} - 1} }}\left( { - \frac{1}{{{x^2}}}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{x\sqrt {{x^2} - 1} }} \cr} $$
