Answer
$$\frac{{dy}}{{ds}} = {2^{{s^2} + 1}}s\left( {\ln 2} \right)$$
Work Step by Step
$$\eqalign{
& y = {2^{\left( {{s^2}} \right)}} \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}s \cr
& \frac{{dy}}{{ds}} = \frac{d}{{ds}}\left[ {{2^{\left( {{s^2}} \right)}}} \right] \cr
& {\text{use the rule }}\frac{d}{{dx}}\left[ {{a^u}} \right] = {a^x}\ln a\frac{{du}}{{dx}}. \cr
& {\text{For this exercise}}\cr
&{\text{let }}a = 2{\text{ and }}u = {s^2},{\text{ }}x = s\cr
&{\text{Then}}{\text{,}} \cr
& \frac{{dy}}{{ds}} = {2^{\left( {{s^2}} \right)}}\ln 2\frac{d}{{ds}}\left[ {{s^2}} \right] \cr
& {\text{solve the derivative and simplify}} \cr
& \frac{{dy}}{{ds}} = {2^{\left( {{s^2}} \right)}}\ln 2\left( {2s} \right) \cr
& \frac{{dy}}{{ds}} = {2^{{s^2} + 1}}s\left( {\ln 2} \right) \cr} $$
