Answer
$$\frac{{dy}}{{ds}} = \frac{s}{{\sqrt {{s^2} - 1} }} - \frac{1}{{\left| s \right|\sqrt {{s^2} - 1} }}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {{s^2} - 1} - {\sec ^{ - 1}}s \cr
& {\text{find the derivative of }}y{\text{ with respect to }}s \cr
& \frac{{dy}}{{ds}} = \frac{{d\left( {\sqrt {{s^2} - 1} } \right)}}{{ds}} + \frac{{d\left( {{{\sec }^{ - 1}}s} \right)}}{{ds}} \cr
& {\text{we can use the formula }}\cr
& \frac{{d\left( {{{\sec }^{ - 1}}u} \right)}}{{dt}} = \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dt}}\,\,\,\,\left| u \right| > 1.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr
& {\text{and using the chain rule}} \cr
& \frac{{dy}}{{ds}} = \frac{1}{2}{\left( {{s^2} - 1} \right)^{ - 1/2}}\left( {2s} \right) - \frac{1}{{\left| s \right|\sqrt {{s^2} - 1} }}\left( 1 \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{ds}} = \frac{s}{{\sqrt {{s^2} - 1} }} - \frac{1}{{\left| s \right|\sqrt {{s^2} - 1} }} \cr} $$
