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: 24

Answer

$$\frac{{dy}}{{dx}} = - 12\cosh 4x{\sinh ^2}4x$$

Work Step by Step

$$\eqalign{ & y = - {\sinh ^3}4x \cr & y = - {\left( {\sinh 4x} \right)^3} \cr & {\text{computing }}dy/dx \cr & \frac{{dy}}{{dx}} = - \frac{d}{{dx}}{\left( {\sinh 4x} \right)^3} \cr & {\text{by the chain rule}} \cr & \frac{{dy}}{{dx}} = - 3{\left( {\sinh 4x} \right)^2}\frac{d}{{dx}}\left( {\sinh 4x} \right) \cr & \frac{{dy}}{{dx}} = - 3{\left( {\sinh 4x} \right)^2}\left( {\cosh 4x} \right)\frac{d}{{dx}}\left( {4x} \right) \cr & \frac{{dy}}{{dx}} = - 3{\left( {\sinh 4x} \right)^2}\left( {\cosh 4x} \right)\left( 4 \right) \cr & {\text{multiplying}} \cr & \frac{{dy}}{{dx}} = - 12\cosh 4x{\sinh ^2}4x \cr} $$
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