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: 59



Work Step by Step

$$y=\ln\Big(\frac{e^\theta}{1+e^{\theta}}\Big)$$ We have $$(\ln \theta)'=\frac{1}{\theta}$$ $$(e^{\theta})'=e^\theta$$ So the derivative of $y$ is $$y'=\frac{1}{\frac{e^{\theta}}{1+e^\theta}}\times\Big(\frac{e^\theta}{1+e^{\theta}}\Big)'=\frac{1+e^{\theta}}{e^{\theta}}\times\frac{e^{\theta}(1+e^\theta)-e^{\theta}(0+e^\theta)}{(1+e^{\theta})^2}$$ $$y'=\frac{1+e^{\theta}}{e^{\theta}}\times\frac{e^\theta+e^{2\theta}-e^{2\theta}}{(1+e^{\theta})^2}=\frac{1+e^{\theta}}{e^{\theta}}\times\frac{e^\theta}{(1+e^{\theta})^2}$$ $$y'=\frac{1}{1+e^{\theta}}$$
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