## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{d\theta }} = \sec \theta$$
\eqalign{ & y = \ln \left( {\sec \theta + \tan \theta } \right) \cr & {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr & \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {\ln \left( {\sec \theta + \tan \theta } \right)} \right] \cr & {\text{use the rule }}\frac{d}{{d\theta }}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{d\theta }} \cr & \frac{{dy}}{{d\theta }} = \frac{1}{{\sec \theta + \tan \theta }}\frac{d}{{d\theta }}\left[ {\sec \theta + \tan \theta } \right] \cr & {\text{solve the derivatives}} \cr & \frac{{dy}}{{d\theta }} = \frac{1}{{\sec \theta + \tan \theta }}\left( {\sec \theta tan\theta + se{c^2}\theta } \right) \cr & {\text{simplify}} \cr & \frac{{dy}}{{d\theta }} = \frac{{\sec \theta tan\theta + se{c^2}\theta }}{{\sec \theta + \tan \theta }} \cr & \frac{{dy}}{{d\theta }} = \frac{{\sec \theta \left( {tan\theta + sec\theta } \right)}}{{\sec \theta + \tan \theta }} \cr & \frac{{dy}}{{d\theta }} = \sec \theta \cr}