Answer
$g'\left( x \right) = - \frac{1}{{2\sqrt x \sqrt {1 - x} }}$
Work Step by Step
$$\eqalign{
& g\left( x \right) = \arccos \sqrt x \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \arccos \sqrt x \cr
& {\text{Use the Derivatives of Inverse Trigonometric Functions }} \cr
& {\text{and the chain rule}}{\text{. }}\frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}u} \right] = - \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}},\;{\text{ }}u = \sqrt x \cr
& {\text{Simplify}} \cr
& g'\left( x \right) = - \frac{1}{{\sqrt {1 - {{\left( {\sqrt x } \right)}^2}} }}\frac{d}{{dx}}\left[ {\sqrt x } \right] \cr
& g'\left( x \right) = - \frac{1}{{\sqrt {1 - x} }}\left( {\frac{1}{{2\sqrt x }}} \right) \cr
& g'\left( x \right) = - \frac{1}{{2\sqrt x \sqrt {1 - x} }} \cr} $$
