Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 23


$$ y'= -sech\, x-sech\, x \, csch^2x.$$

Work Step by Step

Since $ y=sech x\coth x $, then the derivative, by using the product rule $(uv)'=uv'+u'v $ and the facts that $(\ sech \ x)= -\ sech \ x \tanh x $ and $(\coth x)'=-csch^2 x $, is given by $$ y'= -sech\, x\tanh x\coth x-sech\, x \, csch^2x\\ y'= -sech\, x-sech\, x \, csch^2x$$
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