Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 13 - The Trigonometric Functions - 13.2 Derivatives of Trigonometric Functions - 13.2 Exercises - Page 689: 34


$$\frac{d}{{dx}}\left[ {\sec x} \right] = \sec x\tan x$$

Work Step by Step

$$\eqalign{ & \sec x = \frac{1}{{\cos x}} \cr & {\text{differentiate with respect to }}x \cr & \frac{d}{{dx}}\left[ {\sec x} \right] = \frac{d}{{dx}}\left[ {\frac{1}{{\cos x}}} \right] \cr & {\text{by using the quotient rule}} \cr & \frac{d}{{dx}}\left[ {\sec x} \right] = \frac{{\left( {\cos x} \right)\frac{d}{{dx}}\left[ 1 \right] - \frac{d}{{dx}}\left[ {\cos x} \right]}}{{{{\left( {\cos x} \right)}^2}}} \cr & {\text{then}} \cr & \frac{d}{{dx}}\left[ {\sec x} \right] = \frac{{\left( {\cos x} \right)\left( 0 \right) - \left( { - \sin x} \right)}}{{{{\left( {\cos x} \right)}^2}}} \cr & {\text{simplifying}} \cr & \frac{d}{{dx}}\left[ {\sec x} \right] = \frac{{\sin x}}{{{{\left( {\cos x} \right)}^2}}} \cr & \frac{d}{{dx}}\left[ {\sec x} \right] = \left( {\frac{1}{{\cos x}}} \right)\left( {\frac{{\sin x}}{{\cos x}}} \right) \cr & \frac{d}{{dx}}\left[ {\sec x} \right] = \sec x\tan x \cr} $$
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