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: 41

Answer

$y'=\frac{1}{2}\sqrt \frac{x-1}{x^{4}+1}[\frac{1}{x-1}-\frac{4x^{3}}{x^{4}+1}]$ or $y'=\sqrt \frac{x-1}{x^{4}+1}[\frac{1}{2x-2}-\frac{2x^{3}}{x^{4}+1}]$

Work Step by Step

Use logarithmic differentiation to find the derivative of the function $y=\sqrt \frac{x-1}{x^{4}+1}$ Taking the log on both sides: $lny=ln[\sqrt \frac{x-1}{x^{4}+1}]$ Use logarithmic properties $ln(xy)=lnx+lny$ , $ln(\frac{x} {y})=lnx-lny$ and $ln(x^{y})=ylnx$. $lny=\frac{1}{2}[ln(x-1)-ln(x^{4}+1)]$ Differentiate with respect to $x$. $\frac{y'}{y}=\frac{1}{2}[\frac{1}{x-1}-\frac{4x^{3}}{x^{4}+1}]$ $y'=y\frac{1}{2}[\frac{1}{x-1}-\frac{4x^{3}}{x^{4}+1}]$ Hence, $y'=\frac{1}{2}\sqrt \frac{x-1}{x^{4}+1}[\frac{1}{x-1}-\frac{4x^{3}}{x^{4}+1}]$ or $y'=\sqrt \frac{x-1}{x^{4}+1}[\frac{1}{2x-2}-\frac{2x^{3}}{x^{4}+1}]$
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