Answer
$$\frac{{dy}}{{dx}} = \frac{{2x}}{{1 - 4{x^2}}} + {\tanh ^{ - 1}}2x$$
Work Step by Step
$$\eqalign{
& y = x{\tanh ^{ - 1}}2x \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x{{\tanh }^{ - 1}}2x} \right] \cr
& {\text{Use the product rule}} \cr
& \frac{{dy}}{{dx}} = x\frac{d}{{dx}}\left[ {{{\tanh }^{ - 1}}2x} \right] + {\tanh ^{ - 1}}2x\frac{d}{{dx}}\left[ x \right] \cr
& {\text{Using the rules of theorem 5}}{\text{.20 }}\left( {{\text{Page 389}}} \right) \cr
& \frac{{dy}}{{dx}} = x\left( {\frac{1}{{1 - {{\left( {2x} \right)}^2}}}} \right)\frac{d}{{dx}}\left[ {2x} \right] + {\tanh ^{ - 1}}2x\left( 1 \right) \cr
& \frac{{dy}}{{dx}} = x\left( {\frac{1}{{1 - {{\left( {2x} \right)}^2}}}} \right)\left( 2 \right) + {\tanh ^{ - 1}}2x\left( 1 \right) \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{{2x}}{{1 - 4{x^2}}} + {\tanh ^{ - 1}}2x \cr} $$
