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