Answer
$$\frac{{dy}}{{dx}} = \frac{{{e^x}}}{{\left| x \right|\sqrt {{x^2} - 1} }} + {e^x}{\sec ^{ - 1}}x$$
Work Step by Step
$$\eqalign{
& y = {e^x}{\sec ^{ - 1}}x \cr
& {\text{differentiate using the product rule}} \cr
& \frac{{dy}}{{dx}} = {e^x}\left( {{{\sec }^{ - 1}}x} \right)' + {\sec ^{ - 1}}x\left( {{e^x}} \right)' \cr
& {\text{compute derivatives}} \cr
& \frac{{dy}}{{dx}} = {e^x}\left( {\frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}} \right) + {\sec ^{ - 1}}x\left( {{e^x}} \right) \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{\left| x \right|\sqrt {{x^2} - 1} }} + {e^x}{\sec ^{ - 1}}x \cr} $$