Let #color(green)(g(x)=sqrt(x))# and #f(x)=arcsinx#
Then#color(blue)(f(color(green)(g(x)))=arcsinsqrtx)#
Since the given function is a composite function we should differentiate using chain rule.
#color(red)(f(g(x))')=color(red)(f')(color(green)(g(x)))*color(red)(g'(x))#
Let us compute #color(red)(f'(color(green)(g(x)))) and color(red)(g'(x))#
#f(x)=arcsinx#
#f'(x)=1/(sqrt(1-x^2))#
#color(red)(f'(color(green)(g(x)))=1/(sqrt(1-color(green)(g(x))^2))#
#f'(color(green)(g(x)))=1/(sqrt(1-color(green)(sqrtx)^2))#
#color(red)(f'(g(x))=1/(sqrt(1-x)))#
#color(red)(g'(x))=?#
#color(green)(g(x)=sqrtx)#
#color(red)(g'(x)=1/(2sqrtx))#
#color(red)(f(g(x))')=color(red)(f'(g(x)))*color(red)(g'(x))#
#color(red)(f(g(x))')=1/(sqrt(1-x))*1/(2sqrtx)#
#color(red)(f(g(x))')=1/(2sqrt(x(1-x)))#
Therefore,
#color(blue)((arcsinsqrtx)'=1/(2sqrt(x(1-x)))#