Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 42

Answer

$$ g'(t) =\frac{t\sinh^{-1}t}{\sqrt{t^2-1}}+\frac{\sqrt{t^2-1}}{\sqrt{t^2+1}}.$$

Work Step by Step

Since $ g(t)=\sqrt{t^2-1}\sinh^{-1}t $, then the derivative, by using the product rule, is given by $$ g'(t)=\frac{(t^2-1)'}{2\sqrt{t^2-1}}\sinh^{-1}t+\sqrt{t^2-1}\frac{1}{\sqrt{t^2+1}}\\ =\frac{2t\sinh^{-1}t}{2\sqrt{t^2-1}}+\frac{\sqrt{t^2-1}}{\sqrt{t^2+1}}\\ =\frac{t\sinh^{-1}t}{\sqrt{t^2-1}}+\frac{\sqrt{t^2-1}}{\sqrt{t^2+1}}$$
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