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