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: 27

Answer

$$\frac{{dy}}{{dx}} = - \frac{{3{{\operatorname{csch} }^2}3x}}{{2\sqrt {\coth 3x} }}$$

Work Step by Step

$$\eqalign{ & y = \ln \operatorname{sech} 2x \cr & {\text{computing }}dy/dx \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}{\left( {\coth 3x} \right)^{1/2}} \cr & {\text{by the chain rule}} \cr & \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\coth 3x} \right)^{1/2 - 1}}\frac{d}{{dx}}\left( {\coth 3x} \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\coth 3x} \right)^{1/2 - 1}}\left( { - {{\operatorname{csch} }^2}3x} \right)\frac{d}{{dx}}\left( {3x} \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\coth 3x} \right)^{ - 1/2}}\left( { - {{\operatorname{csch} }^2}3x} \right)\left( 3 \right) \cr & {\text{multiplying}} \cr & \frac{{dy}}{{dx}} = - \frac{{3{{\operatorname{csch} }^2}3x{{\left( {\coth 3x} \right)}^{ - 1/2}}}}{2} \cr & \frac{{dy}}{{dx}} = - \frac{{3{{\operatorname{csch} }^2}3x}}{{2{{\left( {\coth 3x} \right)}^{ - 1/2}}}} \cr & \frac{{dy}}{{dx}} = - \frac{{3{{\operatorname{csch} }^2}3x}}{{2\sqrt {\coth 3x} }} \cr} $$
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