Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 223: 12



Work Step by Step

$h(x)=\ln(x+\sqrt{x^2-1})\\ h'(x)=\frac{1}{x+\sqrt{x^2-1}}\times{(1+\frac{x}{\sqrt{x^2-1}})}\\ h'(x)=\frac{1}{x+\sqrt{x^2-1}}\times{\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}}\\ h'(x)=\frac{1}{\sqrt{x^2-1}}$
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