Answer
$$y' = \frac{{{2^{{{\sin }^{ - 1}}x}}\left( {\ln 2} \right)}}{{\sqrt {1 - {x^2}} }}$$
Work Step by Step
$$\eqalign{
& y = {2^{{{\sin }^{ - 1}}x}} \cr
& {\text{find the derivative}} \cr
& y' = \left( {{2^{{{\sin }^{ - 1}}x}}} \right)' \cr
& {\text{differentiate using the chain rule}} \cr
& \frac{d}{{dx}}\left[ {{a^u}} \right] = {a^u}\left( {\ln a} \right)\frac{{du}}{{dx}} \cr
& \cr
& y' = {2^{{{\sin }^{ - 1}}x}}\left( {\ln 2} \right)\left( {{{\sin }^{ - 1}}x} \right)' \cr
& {\text{differentiate using the formula }} \cr
& \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}}{\text{ }}\left( {{\text{see page 467}}} \right) \cr
& y' = {2^{{{\sin }^{ - 1}}x}}\left( {\ln 2} \right)\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) \cr
& {\text{simplifying}}{\text{, }} \cr
& y' = \frac{{{2^{{{\sin }^{ - 1}}x}}\left( {\ln 2} \right)}}{{\sqrt {1 - {x^2}} }} \cr} $$
