## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 223: 26

#### Answer

$y'=\frac{1}{x(1+lnx)}$ and $y''=-\frac{2+lnx}{x^{2}(1+lnx)^{2}}$

#### Work Step by Step

Find the first and second derivative of the function $y=ln(1+lnx)$ $y'=\frac{1}{x(1+lnx)}$ and $y''=-\frac{2+lnx}{x^{2}(1+lnx)^{2}}$ Differentiate the given function with respect to $x$. $y=\frac{d}{dx}ln(1+lnx)$ Thus, $y'=\frac{1}{x(1+lnx)}$ Now, $y''=\frac{d}{dx}\frac{1}{x(1+lnx)}$ $=\frac{1}{x^{2}(1+lnx)^{2}}\frac{d}{dx}\frac{1}{x(1+lnx)}$ $y''=-\frac{2+lnx}{x^{2}(1+lnx)^{2}}$ Hence, $y'=\frac{1}{x(1+lnx)}$ and $y''=-\frac{2+lnx}{x^{2}(1+lnx)^{2}}$

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.