## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 60

#### Answer

$$y'=\frac{1}{2\theta(1+\sqrt\theta)}$$

#### Work Step by Step

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

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