Answer
$$\frac{{dy}}{{dx}} = \frac{{\sqrt {{x^2} - 4} }}{x}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {{x^2} - 4} - 2\operatorname{arcsec} \frac{x}{2},{\text{ }}2 < x < 4 \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sqrt {{x^2} - 4} } \right] - \frac{d}{{dx}}\left[ {2\operatorname{arcsec} \frac{x}{2}} \right] \cr
& {\text{Using rules for differentiation}} \cr
& \frac{{dy}}{{dx}} = \frac{{2x}}{{2\sqrt {{x^2} - 4} }} - 2\left( {\frac{{1/2}}{{\left| {x/2} \right|\sqrt {{{\left( {x/2} \right)}^2} - 1} }}} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{x}{{\sqrt {{x^2} - 4} }} - 2\left( {\frac{2}{{\left| x \right|\sqrt {{x^2} - 4} }}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{x}{{\sqrt {{x^2} - 4} }} - \frac{4}{{\left| x \right|\sqrt {{x^2} - 4} }} \cr
& {\text{Using the fact that }}2 < x < 4 \cr
& \frac{{dy}}{{dx}} = \frac{x}{{\sqrt {{x^2} - 4} }} - \frac{4}{{x\sqrt {{x^2} - 4} }} \cr
& {\text{Add functions}} \cr
& \frac{{dy}}{{dx}} = \frac{{{x^2} - 4}}{{x\sqrt {{x^2} - 4} }} \cr
& \frac{{dy}}{{dx}} = \frac{{\sqrt {{x^2} - 4} }}{x} \cr} $$
