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: 30

Answer

$$\frac{{dy}}{{dx}} = \frac{{1 + x\coth x}}{{\operatorname{csch} x}}$$

Work Step by Step

$$\eqalign{ & y = x/\operatorname{csch} x \cr & {\text{computing }}dy/dx \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {x/\operatorname{csch} x} \right) \cr & {\text{by the quotient rule}} \cr & \frac{{dy}}{{dx}} = \frac{{\operatorname{csch} x\left( x \right)' - x\left( {\operatorname{csch} x} \right)'}}{{{{\left( {\operatorname{csch} x} \right)}^2}}} \cr & {\text{using basic formulas for differentiation}} \cr & \frac{{dy}}{{dx}} = \frac{{\operatorname{csch} x\left( 1 \right) - x\left( { - \operatorname{csch} x\coth x} \right)}}{{{{\left( {\operatorname{csch} x} \right)}^2}}} \cr & {\text{multiplying}} \cr & \frac{{dy}}{{dx}} = \frac{{\operatorname{csch} x + xcschx\coth x}}{{{{\operatorname{csch} }^2}x}} \cr & {\text{Simplify}} \cr & \frac{{dy}}{{dx}} = \frac{{1 + x\coth x}}{{\operatorname{csch} x}} \cr} $$
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