Answer
$$\frac{{dy}}{{dx}} = \frac{1}{x} + \frac{{{{\sec }^{ - 1}}\sqrt x }}{{\sqrt {x - 1} }}$$
Work Step by Step
$$\eqalign{
& y = 2\sqrt {x - 1} {\sec ^{ - 1}}\sqrt x \cr
& {\text{find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{{d\left( {2\sqrt {x - 1} {{\sec }^{ - 1}}\sqrt x } \right)}}{{dx}} \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{dx}} = 2\sqrt {x - 1} \frac{{d\left( {{{\sec }^{ - 1}}\sqrt x } \right)}}{{dx}} + {\sec ^{ - 1}}\sqrt x \frac{{d\left( {2\sqrt {x - 1} } \right)}}{{dx}} \cr
& {\text{use the formula }}\cr
&\frac{{d\left( {{{\sec }^{ - 1}}u} \right)}}{{dx}} = \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}}\cr
&{\text{ and the chain rule}} \cr
& \frac{{dy}}{{dx}} = 2\sqrt {x - 1} \left( {\frac{1}{{\left| {\sqrt x } \right|\sqrt {{{\left( {\sqrt x } \right)}^2} - 1} }}} \right)\frac{d}{{dx}}\left( {\sqrt x } \right) + 2{\sec ^{ - 1}}\sqrt x \left( {\frac{1}{{2\sqrt {x - 1} }}} \right) \cr
& \frac{{dy}}{{dx}} = 2\sqrt {x - 1} \left( {\frac{1}{{\sqrt x \sqrt {x - 1} }}} \right)\left( {\frac{1}{{2\sqrt x }}} \right) + 2{\sec ^{ - 1}}\sqrt x \left( {\frac{1}{{2\sqrt {x - 1} }}} \right) \cr
& {\text{simplifying, we get:}} \cr
& \frac{{dy}}{{dx}} = 2\sqrt {x - 1} \left( {\frac{1}{{2x\sqrt {x - 1} }}} \right) + \frac{{2{{\sec }^{ - 1}}\sqrt x }}{{2\sqrt {x - 1} }} \cr
& \frac{{dy}}{{dx}} = \frac{1}{x} + \frac{{{{\sec }^{ - 1}}\sqrt x }}{{\sqrt {x - 1} }} \cr} $$
