Answer
$$\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x \left( {1 - x} \right)}}$$
Work Step by Step
$$\eqalign{
& y = {\tanh ^{ - 1}}\sqrt x \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\tanh }^{ - 1}}\sqrt x } \right] \cr
& {\text{Use Theorem 5}}{\text{.20 }}\frac{d}{{dx}}\left[ {{{\tanh }^{ - 1}}u} \right] = \frac{{u'}}{{1 - {u^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dx}}\left[ {\sqrt x } \right]}}{{1 - {{\left( {\sqrt x } \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{1}{{2\sqrt x }}}}{{1 - x}} \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x \left( {1 - x} \right)}} \cr} $$
