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: 21

Answer

$$ - \operatorname{csch} x\coth x$$

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}\left( {\operatorname{csch} x} \right) = - \operatorname{csch} x\coth x \cr & {\text{ hyperbolic functions for cosecant}} \cr & \frac{d}{{dx}}\left( {\operatorname{csch} x} \right) = \frac{d}{{dx}}\left( {\frac{1}{{\sinh x}}} \right) \cr & {\text{differentiate by the product rule}} \cr & \frac{{\sinh x\frac{d}{{dx}}\left[ 1 \right] - 1\frac{d}{{dx}}\left[ {\sinh x} \right]}}{{{{\left( {\sinh x} \right)}^2}}} \cr & \frac{{\sinh x\left( 0 \right) - \cosh x}}{{{{\left( {\sinh x} \right)}^2}}} \cr & {\text{simplify}} \cr & \frac{{ - \sinh x}}{{{{\left( {\sinh x} \right)}^2}}} \cr & or \cr & - \frac{1}{{\sinh x}}\frac{{\cosh x}}{{\sinh x}} \cr & {\text{hyperbolic functions}} \cr & - \operatorname{csch} x\coth x \cr} $$
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