Answer
$$\frac{{dy}}{{dx}} = 2\sec 2x$$
Work Step by Step
$$\eqalign{
& y = {\operatorname{sech} ^{ - 1}}\left( {\cos 2x} \right),{\text{ }}0 < x < \pi /4 \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\operatorname{sech} }^{ - 1}}\left( {\cos 2x} \right)} \right] \cr
& {\text{Use Theorem 5}}{\text{.20 }}\frac{d}{{dx}}\left[ {{{\operatorname{sech} }^{ - 1}}u} \right] = - \frac{{u'}}{{u\sqrt {1 - {u^2}} }} \cr
& \frac{{dy}}{{dx}} = - \frac{{\frac{d}{{dx}}\left[ {\cos 2x} \right]}}{{\cos 2x\sqrt {1 - {{\left( {\cos 2x} \right)}^2}} }} \cr
& \frac{{dy}}{{dx}} = - \frac{{ - 2\sin 2x}}{{\cos 2x\sqrt {1 - {{\cos }^2}2x} }} \cr
& {\text{Using the identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \cr
& \frac{{dy}}{{dx}} = \frac{{2\sin 2x}}{{\cos 2x\sqrt {{{\sin }^2}2x} }} \cr
& \frac{{dy}}{{dx}} = \frac{{2\sin 2x}}{{\cos 2x\sin 2x}} \cr
& \frac{{dy}}{{dx}} = \frac{2}{{\cos 2x}} \cr
& \frac{{dy}}{{dx}} = 2\sec 2x \cr} $$
