Answer
$\frac{{dy}}{{d\theta }} = \sec \theta $
Work Step by Step
$$\eqalign{
& y = {\cosh ^{ - 1}}\left( {\sec \theta } \right),{\text{ 0}} \leqslant \theta < \frac{\pi }{2} \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{{\cosh }^{ - 1}}\left( {\sec \theta } \right)} \right] \cr
& {\text{Use Derivatives of Inverse Hyperbolic Functions }} \cr
& \frac{d}{{d\theta }}\left[ {{{\cos }^{ - 1}}u} \right] = \frac{1}{{\sqrt {{u^2} - 1} }}\frac{{du}}{{d\theta }},{\text{ let }}u = \sec \theta ,{\text{ so}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\sqrt {{{\left( {\sec \theta } \right)}^2} - 1} }}\frac{d}{{d\theta }}\left[ {\sec \theta } \right] \cr
& {\text{Compute the derivative and simplify}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\sqrt {{{\sec }^2}\theta - 1} }}\left( {\sec \theta \tan \theta } \right) \cr
& {\text{Use the identity }}1 + {\tan ^2}\theta = {\sec ^2}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\sqrt {{{\tan }^2}\theta } }}\left( {\sec \theta \tan \theta } \right) \cr
& \frac{{dy}}{{d\theta }} = \sec \theta \cr} $$
