Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises: 23

Answer

$$\frac{{dy}}{{dx}} = \sinh 2x$$

Work Step by Step

$$\eqalign{ & y = {\cosh ^2}x \cr & y = {\left( {\cosh x} \right)^2} \cr & {\text{computing }}dy/dx \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}{\left( {\cosh x} \right)^2} \cr & {\text{by the chain rule}} \cr & \frac{{dy}}{{dx}} = 2{\left( {\cosh x} \right)^{2 - 1}}\frac{d}{{dx}}\left( {\cosh x} \right) \cr & \frac{{dy}}{{dx}} = 2\cosh x\left( {\sinh x} \right) \cr & {\text{multiplying}} \cr & \frac{{dy}}{{dx}} = 2\sinh x\cosh x \cr & or \cr & \frac{{dy}}{{dx}} = \sinh 2x \cr} $$
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