Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.8 Hyperbolic Functions And Hanging Cables - Exercises Set 6.8 - Page 481: 13


$$y' = \frac{1}{{{x^2}}}\operatorname{csch} \left( {\frac{1}{x}} \right)\coth \left( {\frac{1}{x}} \right)$$

Work Step by Step

$$\eqalign{ & y = \operatorname{csch} \left( {1/x} \right) \cr & {\text{find the derivatvive}} \cr & y' = \left[ {\operatorname{csch} \left( {\frac{1}{x}} \right)} \right]' \cr & {\text{use the theorem 6}}{\text{.8}}{\text{.3 }}\left( {{\text{ see page 476}}} \right) \cr & \frac{d}{{dx}}\left[ {\operatorname{csch} \left( {\frac{1}{x}} \right)} \right] = - \operatorname{csch} u\coth u\frac{{du}}{{dx}} \cr & y' = - \operatorname{csch} \left( {\frac{1}{x}} \right)\coth \left( {\frac{1}{x}} \right)\left( {\frac{1}{x}} \right)' \cr & {\text{compute derivative}} \cr & y' = - \operatorname{csch} \left( {\frac{1}{x}} \right)\coth \left( {\frac{1}{x}} \right)\left( { - \frac{1}{{{x^2}}}} \right) \cr & {\text{simplifying}} \cr & y' = \frac{1}{{{x^2}}}\operatorname{csch} \left( {\frac{1}{x}} \right)\coth \left( {\frac{1}{x}} \right) \cr} $$
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