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: 35


$$y'=\frac{2}{t(1-\ln t)^2}$$

Work Step by Step

$$y=\frac{1+\ln t}{1-\ln t}$$ We have $$y'=\frac{(1+\ln t)'(1-\ln t)-(1+\ln t)(1-\ln t)'}{(1-\ln t)^2}$$ $$y'=\frac{\frac{1}{t}(1-\ln t)-(1+\ln t)\Big(-\frac{1}{t}\Big)}{(1-\ln t)^2}$$ $$y'=\frac{\frac{1}{t}(1-\ln t)+\frac{1}{t}(1+\ln t)}{(1-\ln t)^2}$$ $$y'=\frac{\frac{1}{t}(1-\ln t+1+\ln t)}{(1-\ln t)^2}$$ $$y'=\frac{\frac{1}{t}(2)}{(1-\ln t)^2}$$ $$y'=\frac{2}{t(1-\ln t)^2}$$
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