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: 19

Answer

$$ - {\operatorname{csch} ^2}x$$

Work Step by Step

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