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 - Page 223: 24


$y^{\prime \prime}=-\frac{3+\ln x}{x^{2}(\ln x+1)^{3}}$

Work Step by Step

$y=\frac{\ln x}{\ln x+1}\\ y^{\prime}=\frac{(\ln x+1)\left(\frac{1}{x}\right)-(\ln x)\left(\frac{1}{x}\right)}{(\ln x+1)^{2}}\\ y^{\prime}=\frac{\left(\frac{\ln x}{x}\right)+\left(\frac{1}{x}\right)-\left(\frac{\ln x}{x}\right)}{(\ln x+1)^{2}}\\ y^{\prime}=\frac{1}{x(\ln x+1)^{2}}\\ y^{\prime \prime}=\frac{\left(x(\ln x+1)^{2}\right)(0)-((1)((x)(2(\ln x+1)\left(\frac{1}{x}\right)))+(\ln x+1)^{2}(1))}{\left(x(\ln x+1)^{2}\right)^2}\\ y^{\prime \prime}=\frac{-2\left(\ln x+1\right)-\left(\ln x+1\right)^2}{x^2(\ln x+1)^{4}}\\ y^{\prime \prime}=\frac{\left(\ln x+1\right)\left(-2-(\ln x+1)\right)}{x^2(\ln x+1)^{4}}\\ y^{\prime \prime}=\frac{-2-\ln x-1}{x^{2}(\ln x+1)^{3}}\\ y^{\prime \prime}=-\frac{3+\ln x}{x^{2}(\ln x+1)^{3}}$
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