Answer
$$\frac{{dy}}{{ds}} = \frac{1}{{\left| s \right|\sqrt {25{s^2} - 1} }}$$
Work Step by Step
$$\eqalign{
& y = {\sec ^{ - 1}}5s \cr
& {\text{find the derivative of }}y{\text{ with respect to }}s \cr
& \frac{{dy}}{{ds}} = \frac{{d\left( {{{\sec }^{ - 1}}5s} \right)}}{{ds}} \cr
& {\text{we can use the formula }}\cr
& \frac{{d\left( {{{\sec }^{ - 1}}u} \right)}}{{ds}} = \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{ds}}\,\,\,\,\left| u \right| > 1.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr
& {\text{here }}u = 5s,\,\,{\text{then}} \cr
& \frac{{dy}}{{ds}} = \frac{1}{{\left| {5s} \right|\sqrt {{{\left( {5s} \right)}^2} - 1} }}\frac{{d\left( {5s} \right)}}{{ds}} \cr
& \frac{{dy}}{{ds}} = \frac{1}{{\left| {5s} \right|\sqrt {25{s^2} - 1} }}\left( 5 \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{ds}} = \frac{1}{{\left| s \right|\sqrt {25{s^2} - 1} }} \cr} $$
