University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 199: 34



Work Step by Step

We evaluate the function: $y=\ln\frac{(x+1)}{\sqrt{(x-1)}}$ $y=\ln{(x+1)}-ln{(\sqrt{(x-1)})}$ $y=\ln{(x+1)}-\frac{1}{2}ln{{(x-1)}}$ on differentiating the above: $\frac{dy}{dx}=\frac{d({\ln{(x+1)}-\frac{1}{2}ln{{(x-1)}}})}{dx}$ $\frac{dy}{dx}=\frac{1}{(1+x)}\frac{d(1+x)}{dx}-\frac{1}{2}\frac{1}{x-1}\frac{d(x-1)}{dx}$ or $\frac{dy}{dx}=\frac{1}{1+x}-\frac{1}{2(x-1)}$ or $\frac{dy}{dx}=\frac{2x-2-1-x}{2(x^2-1)}$ or ${dy}=(\frac{x-3}{2(x^2-1)}){dx}$
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