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: 50

Answer

$$f'\left( x \right) = \frac{{\left| x \right|}}{{x\sqrt {{x^2} + 4} }}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {\operatorname{csch} ^{ - 1}}\left( {2/x} \right) \cr & {\text{find the derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {{{\operatorname{csch} }^{ - 1}}\left( {2/x} \right)} \right) \cr & {\text{use derivatives of the inverse hyperbolic functions}} \cr & f'\left( x \right) = - \frac{1}{{\left| {2/x} \right|\sqrt {1 + {{\left( {2/x} \right)}^2}} }}\frac{d}{{dx}}\left( {\frac{2}{x}} \right) \cr & f'\left( x \right) = - \frac{1}{{\left| {2/x} \right|\sqrt {1 + {{\left( {2/x} \right)}^2}} }}\left( { - \frac{2}{{{x^2}}}} \right) \cr & {\text{simplify}} \cr & f'\left( x \right) = \frac{{\left| x \right|}}{{2\sqrt {\frac{{{x^2} + 4}}{{{x^2}}}} }}\left( {\frac{2}{{{x^2}}}} \right) \cr & f'\left( x \right) = \frac{{\left| x \right|}}{{\frac{{\sqrt {{x^2} + 4} }}{x}}}\left( {\frac{1}{{{x^2}}}} \right) \cr & f'\left( x \right) = \frac{{\left| x \right|}}{{x\sqrt {{x^2} + 4} }} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.