Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises: 12

Answer

$\frac{1}{\sqrt{x^2-1}}$

Work Step by Step

$\frac{d}{dx}\ln(x+\sqrt{x^2-1})$ $=\frac{1}{x+\sqrt{x^2-1}}*\frac{d}{dx}(x+\sqrt{x^2-1})$ $=\frac{1}{x+\sqrt{x^2-1}}*\frac{d}{dx}(x+(x^2-1)^\frac{1}{2})$ $=\frac{1}{x+\sqrt{x^2-1}}*(1+\frac{1}{2}(x^2-1)^{-\frac{1}{2}}\frac{d}{dx}(x^2-1))$ $=\frac{1}{x+\sqrt{x^2-1}}*(1+\frac{1}{2\sqrt{x^2-1}}*2x)$ $=\frac{1}{x+\sqrt{x^2-1}}*(1+\frac{x}{\sqrt{x^2-1}})$ $=\frac{1}{x+\sqrt{x^2-1}}*(\frac{\sqrt{x^2-1}}{\sqrt{x^2-1}}+\frac{x}{\sqrt{x^2-1}})$ $=\frac{1}{x+\sqrt{x^2-1}}*\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}$ $=\frac{1}{\sqrt{x^2-1}}$
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