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: 51

Answer

$$f'\left( x \right) = {\sinh ^{ - 1}}x$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = x{\sinh ^{ - 1}}x - \sqrt {{x^2} + 1} \cr & {\text{find the derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {x{{\sinh }^{ - 1}}x - \sqrt {{x^2} + 1} } \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {x{{\sinh }^{ - 1}}x} \right) - \frac{d}{{dx}}\left( {\sqrt {{x^2} + 1} } \right) \cr & {\text{by the product rule}} \cr & f'\left( x \right) = x\frac{d}{{dx}}\left( {{{\sinh }^{ - 1}}x} \right) + {\sinh ^{ - 1}}x\frac{d}{{dx}}\left( x \right) - \frac{d}{{dx}}\left( {\sqrt {{x^2} + 1} } \right) \cr & {\text{use derivatives of the inverse hyperbolic functions and the chain rule}} \cr & f'\left( x \right) = x\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right) + {\sinh ^{ - 1}}x\left( 1 \right) - \frac{{2x}}{{2\sqrt {{x^2} + 1} }} \cr & {\text{simplify}} \cr & f'\left( x \right) = \frac{x}{{\sqrt {{x^2} + 1} }} + {\sinh ^{ - 1}}x - \frac{x}{{\sqrt {{x^2} + 1} }} \cr & f'\left( x \right) = {\sinh ^{ - 1}}x \cr} $$
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