Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises - Page 325: 49


$\displaystyle \frac{2x^{2}-1}{x(x^{2}-1)}$

Work Step by Step

By theorem 5.3/2: $\displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$ --------------- Here, $u(x)=x\sqrt{x^{2}-1}= \sqrt{x^{2}(x^{2}-1)}=\sqrt{x^{4}-x^{2}}=(x^{4}-x^{2})^{1/2}$ $\displaystyle \frac{du}{dx}$=...chain rule...=$\displaystyle \frac{1}{2}(x^{4}-x^{2})^{-1/2}\cdot(4x^{3}-2x)$ $=\displaystyle \frac{2x(2x^{2}-1)}{2\sqrt{x^{4}-x^{2}}}=\frac{ x(2x^{2}-1 )}{ x\sqrt{x^{2}-1}}=\frac{ 2x^{2}-1 }{\sqrt{x^{2}-1}}$ So, $\displaystyle \frac{d}{dx}[\ln(x\sqrt{x^{2}-1})]=\frac{1}{x\sqrt{x^{2}-1}}\cdot\frac{ 2x^{2}-1 }{\sqrt{x^{2}-1}}=\frac{2x^{2}-1}{x(x^{2}-1)}$
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.