Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 391: 63

Answer

$$\frac{{dy}}{{d\theta }} = {7^{\sec \theta }}{\left( {\ln 7} \right)^2}\sec \theta \tan \theta $$

Work Step by Step

$$\eqalign{ & y = {7^{\sec \theta }}\ln 7 \cr & {\text{find the derivative of }}y{\text{ with respect to }}\theta \cr & \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{7^{\sec \theta }}\ln 7} \right] \cr & \frac{{dy}}{{d\theta }} = \ln 7\frac{d}{{d\theta }}\left[ {{7^{\sec \theta }}} \right] \cr & {\text{Use the general power rule for differentiation }}\cr & \frac{d}{{d\theta }}\left[ {{a^u}} \right] = {a^u}\left( {\ln a} \right)\frac{{du}}{{d\theta }}{\text{ }} \cr & {\text{for this exercise let }}a = 7{\text{ and }}u = \sec \theta {\text{; then}}{\text{,}} \cr & \frac{{dy}}{{d\theta }} = \ln 7\left( {{7^{\sec \theta }}} \right)\left( {\ln 7} \right)\frac{d}{{d\theta }}\left[ {\sec \theta } \right] \cr & {\text{solve the derivative}} \cr & \frac{{dy}}{{d\theta }} = \ln 7\left( {{7^{\sec \theta }}} \right)\left( {\ln 7} \right)\left( {\sec \theta \tan \theta } \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{d\theta }} = {7^{\sec \theta }}{\left( {\ln 7} \right)^2}\sec \theta \tan \theta \cr} $$
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