Answer
$$y' = - \frac{{{e^x}}}{{2x\sqrt {1 - x} }} + {e^x}{\operatorname{sech} ^{ - 1}}\sqrt x $$
Work Step by Step
$$\eqalign{
& y = {e^x}{\operatorname{sech} ^{ - 1}}\sqrt x \cr
& {\text{find the derivatvive}} \cr
& y' = \left( {{e^x}{{\operatorname{sech} }^{ - 1}}\sqrt x } \right)' \cr
& {\text{product rule}} \cr
& y' = {e^x}\left( {{{\operatorname{sech} }^{ - 1}}\sqrt x } \right)' + {\operatorname{sech} ^{ - 1}}\sqrt x \left( {{e^x}} \right)' \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.5 }}\left( {{\text{ see page 479}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {{{\operatorname{sech} }^{ - 1}}u} \right] = - \frac{1}{{u\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}} \cr
& y' = {e^x}\left( { - \frac{1}{{\sqrt x \sqrt {1 - {{\left( {\sqrt x } \right)}^2}} }}} \right)\left( {\sqrt x } \right)' + {\operatorname{sech} ^{ - 1}}\sqrt x \left( {{e^x}} \right) \cr
& y' = {e^x}\left( { - \frac{1}{{\sqrt x \sqrt {1 - x} }}} \right)\left( {\frac{1}{{2\sqrt x }}} \right) + {e^x}{\operatorname{sech} ^{ - 1}}\sqrt x \cr
& {\text{simplifying}} \cr
& y' = {e^x}\left( { - \frac{1}{{2x\sqrt {1 - x} }}} \right) + {e^x}{\operatorname{sech} ^{ - 1}}\sqrt x \cr
& y' = - \frac{{{e^x}}}{{2x\sqrt {1 - x} }} + {e^x}{\operatorname{sech} ^{ - 1}}\sqrt x \cr} $$
