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

Published by Pearson

# Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 45

#### Answer

$$y'=\frac{\Big(2(\cot\theta)(\theta+3)+1\Big)(\sin\theta)\sqrt{\theta+3}}{2(\theta+3)}$$

#### Work Step by Step

$$y=(\sin\theta)\sqrt{\theta+3}$$ 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: $$\ln y=\ln\Big((\sin\theta)\sqrt{\theta+3}\Big)$$ Recall that $\ln(A\times B)=\ln A+\ln B$: $$\ln y=\ln\sin\theta+\ln\sqrt{\theta+3}$$ 2) Then we take the derivative of both sides with respect to $x$ and remember that $(\ln x)'=1/x$: $$\frac{1}{y}\times y'=\frac{1}{\sin\theta}(\sin\theta)'+\frac{1}{\sqrt{\theta+3}}(\sqrt{\theta+3})'$$ $$\frac{y'}{y}=\frac{\cos\theta}{\sin\theta}+\frac{1}{\sqrt{\theta+3}}\times\frac{1}{2\sqrt{\theta+3}}(\theta+3)'$$ $$\frac{y'}{y}=\cot\theta+\frac{1}{2(\theta+3)}$$ $$\frac{y'}{y}=\frac{2(\cot\theta)(\theta+3)+1}{2(\theta+3)}$$ 3) Solve for $y'$: $$y'=\frac{2(\cot\theta)(\theta+3)+1}{2(\theta+3)}\times y$$ 4) Finally, substitute for $y=(\sin\theta)\sqrt{\theta+3}$ $$y'=\frac{\Big(2(\cot\theta)(\theta+3)+1\Big)(\sin\theta)\sqrt{\theta+3}}{2(\theta+3)}$$

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