Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.2 Derivatives And Integrals Involving Logarithmic Functions - Exercises Set 6.2 - Page 425: 30

Answer

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

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}\left[ {\ln \sqrt {\frac{{x - 1}}{{x + 1}}} } \right] \cr & {\text{radical properties}} \cr & \frac{d}{{dx}}\left[ {\ln \frac{{\sqrt {x - 1} }}{{\sqrt {x + 1} }}} \right] \cr & \frac{d}{{dx}}\left[ {\ln \frac{{{{\left( {x - 1} \right)}^{1/2}}}}{{{{\left( {x + 1} \right)}^{1/2}}}}} \right] \cr & {\text{logarithm of a quotient}} \cr & = \frac{d}{{dx}}\left[ {\ln {{\left( {x - 1} \right)}^{1/2}} - \ln {{\left( {x + 1} \right)}^{1/2}}} \right] \cr & {\text{power rule for logarithms}} \cr & = \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {x - 1} \right) - \frac{1}{2}\ln \left( {x + 1} \right)} \right] \cr & {\text{differentiate}} \cr & \frac{1}{2}\left( {\frac{1}{{x - 1}}} \right) - \frac{1}{2}\left( {\frac{1}{{x + 1}}} \right) \cr & {\text{simplify}} \cr & = \frac{1}{2}\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right) \cr & = \frac{1}{2}\left( {\frac{{x + 1 - x + 1}}{{{x^2} - 1}}} \right) \cr & = \frac{1}{2}\left( {\frac{2}{{{x^2} - 1}}} \right) \cr & = \frac{1}{{{x^2} - 1}} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.