## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{d\theta }} = \frac{{tan\left( {\ln \theta } \right)}}{\theta }$$
\eqalign{ & y = \ln \left( {\sec \left( {\ln \theta } \right)} \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 \left( {\ln \theta } \right)} \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 \left( {\ln \theta } \right)}}\frac{d}{{d\theta }}\left[ {\sec \left( {\ln \theta } \right)} \right] \cr & {\text{use the rule }}\frac{d}{{d\theta }}\left[ {\sec u} \right] = \sec u\tan u\frac{{du}}{{d\theta }} \cr & \frac{{dy}}{{d\theta }} = \frac{1}{{\sec \left( {\ln \theta } \right)}}\sec \left( {\ln \theta } \right)tan\left( {\ln \theta } \right)\frac{d}{{d\theta }}\left[ {\ln \theta } \right] \cr & {\text{solve the derivative}} \cr & \frac{{dy}}{{d\theta }} = tan\left( {\ln \theta } \right)\left( {\frac{1}{\theta }} \right) \cr & {\text{simplify}} \cr & \frac{{dy}}{{d\theta }} = \frac{{tan\left( {\ln \theta } \right)}}{\theta } \cr}