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