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: 28

Answer

$$\frac{{dy}}{{dx}} = x{\operatorname{sech} ^2}x + \tanh x$$

Work Step by Step

$$\eqalign{ & y = x\tanh x \cr & {\text{computing }}dy/dx \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {x\tanh x} \right) \cr & {\text{by the product rule}} \cr & \frac{{dy}}{{dx}} = x\frac{d}{{dx}}\left( {\tanh x} \right) + \tanh x\frac{d}{{dx}}\left( x \right) \cr & {\text{using basic formulas for differentiation}} \cr & \frac{{dy}}{{dx}} = x\left( {{{\operatorname{sech} }^2}x} \right) + \tanh x\left( 1 \right) \cr & {\text{multiplying}} \cr & \frac{{dy}}{{dx}} = x{\operatorname{sech} ^2}x + \tanh x \cr} $$
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