University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 27


$$y'=\frac{1}{x(1+\ln x)^2}$$

Work Step by Step

$$y=\frac{\ln x}{1+\ln x}$$ Recall the following Derivative Rules: $$\frac{d}{dt}(\ln t)=\frac{1}{t}$$ $$\frac{d}{dt}\Big(\frac{u}{v}\Big)=\frac{v\frac{du}{dt}-u\frac{dv}{dt}}{v^2}$$ Therefore, we have $$y'=\frac{(\ln x)'(1+\ln x)-(\ln x)(1+\ln x)'}{(1+\ln x)^2}=\frac{\frac{1}{x}(1+\ln x)-\ln x\times\frac{1}{x}}{(1+\ln x)^2}$$ $$y'=\frac{\frac{1}{x}+\frac{1}{x}\ln x-\frac{1}{x}\ln x}{(1+\ln x)^2}=\frac{\frac{1}{x}}{(1+\ln x)^2}$$ $$y'=\frac{1}{x(1+\ln x)^2}$$
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