Let color(green)(g(x)=sqrt(x)) and f(x)=arcsinx
Thencolor(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)))
