Answer
$$\frac{{dy}}{{dx}} = - \frac{2}{{\left| x \right|\sqrt {{x^2} - 4} }}$$
Work Step by Step
$$\eqalign{
& y = {\csc ^{ - 1}}\left( {\frac{x}{2}} \right) \cr
& {\text{find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{{d\left( {{{\csc }^{ - 1}}\left( {\frac{x}{2}} \right)} \right)}}{{dx}} \cr
& {\text{we can use the formula }}\cr
& \frac{{d\left( {{{\csc }^{ - 1}}u} \right)}}{{dx}} = - \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}}.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr
& {\text{here }}u = {x^2} + 1,\,\,{\text{then}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\left| {x/2} \right|\sqrt {{{\left( {x/2} \right)}^2} - 1} }}\frac{{d\left( {x/2} \right)}}{{dx}} \cr
& \frac{{dy}}{{dx}} = - \frac{2}{{\left| x \right|\sqrt {\frac{{{x^2}}}{4} - 1} }}\left( {\frac{1}{2}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\left| x \right|\sqrt {\frac{{{x^2}}}{4} - 1} }} \cr
& \frac{{dy}}{{dx}} = - \frac{2}{{\left| x \right|\sqrt {{x^2} - 4} }} \cr} $$
