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