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: 49

Answer

$$f'\left( v \right) = \frac{{2v}}{{\sqrt {{v^4} + 1} }}$$

Work Step by Step

$$\eqalign{ & f\left( v \right) = {\sinh ^{ - 1}}{v^2} \cr & {\text{find the derivative}} \cr & f'\left( v \right) = \frac{d}{{dv}}\left( {{{\sinh }^{ - 1}}{v^2}} \right) \cr & {\text{use derivatives of the inverse hyperbolic functions}} \cr & f'\left( v \right) = \frac{1}{{\sqrt {{{\left( {{v^2}} \right)}^2} + 1} }}\frac{d}{{dv}}\left( {{v^2}} \right) \cr & f'\left( v \right) = \frac{1}{{\sqrt {{v^4} + 1} }}\left( {2v} \right) \cr & {\text{simplify}} \cr & f'\left( v \right) = \frac{{2v}}{{\sqrt {{v^4} + 1} }} \cr} $$
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