Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2* The Natural Logarithmic Functions - 6.2* Exercises: 38


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

Work Step by Step

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