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

Published by Pearson

# Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 50

#### Answer

$$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\sec\theta}$$

#### Work Step by Step

$$y=\frac{\theta\sin\theta}{\sqrt{\sec\theta}}$$ Logarithmic differentiation helps to simplify the differentiation of otherwise complex functions by using the algebraic properties of logarithm. In detail, 1) Take the natural logarithm of both sides and simplify: $$\ln y=\ln\Big(\frac{\theta\sin\theta}{\sqrt{\sec\theta}}\Big)$$ Recall that $\ln(A\times B)=\ln A+\ln B$ and $\ln(A/B)=\ln A-\ln B$: $$\ln y=\ln(\theta\sin\theta)-\ln(\sqrt{\sec\theta})$$ $$\ln y=\ln\theta+\ln\sin\theta-\ln\sqrt{\sec\theta}$$ 2) Then we take the derivative of both sides with respect to $\theta$ and remember that $(\ln \theta)'=1/\theta$: $$\frac{1}{y}\times y'=\frac{1}{\theta}+\frac{1}{\sin\theta}(\sin\theta)'-\frac{1}{\sqrt{\sec\theta}}(\sqrt{\sec\theta})'$$ $$\frac{y'}{y}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sqrt{\sec\theta}}\times\frac{1}{2\sqrt{\sec\theta}}(\sec\theta)'$$ $$\frac{y'}{y}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{(\sec\theta)'}{2\sec\theta}$$ $$\frac{y'}{y}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{\sec\theta\tan\theta}{2\sec\theta}=\frac{1}{\theta}+\frac{\cos\theta}{\sin\theta}-\frac{\tan\theta}{2}$$ $$\frac{y'}{y}=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\theta\sin\theta}$$ 3) Solve for $y'$: $$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\theta\sin\theta}\times y$$ 4) Finally, substitute for $y=\frac{\theta\sin\theta}{\sqrt{\sec\theta}}$ $$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\theta\sin\theta}\times\frac{\theta\sin\theta}{\sqrt{\sec\theta}}$$ $$y'=\frac{2\sin\theta+2\theta\cos\theta-\theta\sin\theta\tan\theta}{2\sec\theta}$$

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