Answer
$$\frac{{dy}}{{dx}} = \frac{2}{{4 - {x^2}}}$$
Work Step by Step
$$\eqalign{
& y = {\tanh ^{ - 1}}\left( {\frac{x}{2}} \right) \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\tanh }^{ - 1}}\left( {\frac{x}{2}} \right)} \right] \cr
& {\text{Use Theorem 5}}{\text{.20 }}\frac{d}{{dx}}\left[ {{{\tanh }^{ - 1}}u} \right] = \frac{{u'}}{{1 - {u^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dx}}\left[ {x/2} \right]}}{{1 - {{\left( {x/2} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{1}{2}}}{{1 - \frac{{{x^2}}}{4}}} \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{1}{2}}}{{\frac{{4 - {x^2}}}{4}}} \cr
& \frac{{dy}}{{dx}} = \frac{2}{{4 - {x^2}}} \cr} $$
