Answer
$\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{x^2} - 1} }} + {\sec ^{ - 1}}x$
Work Step by Step
$$\eqalign{
& y = x{\sec ^{ - 1}}x \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x{{\sec }^{ - 1}}x} \right] \cr
& {\text{Use the product rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = x\frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}x} \right] + {\sec ^{ - 1}}x\frac{d}{{dx}}\left[ x \right] \cr
& {\text{Where }}\frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}x} \right] = \frac{1}{{x\sqrt {{x^2} - 1} }}{\text{ }},{\text{ so}} \cr
& \frac{{dy}}{{dx}} = x\left( {\frac{1}{{x\sqrt {{x^2} - 1} }}} \right) + {\sec ^{ - 1}}x\left( 1 \right) \cr
& {\text{Multiply and simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{x}{{x\sqrt {{x^2} - 1} }} + {\sec ^{ - 1}}x \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{x^2} - 1} }} + {\sec ^{ - 1}}x \cr} $$
