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

$$y'=\frac{(\theta+5)\tan\theta-5}{\theta\cos\theta}$$
$$y=\frac{\theta+5}{\theta\cos\theta}$$ 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: $$\ln y=\ln\Big(\frac{\theta+5}{\theta\cos\theta}\Big)$$ Recall that $\ln(A\times B)=\ln A+\ln B$ and $\ln(A/B)=\ln A-\ln B$: $$\ln y=\ln(\theta+5)-\ln(\theta\cos\theta)$$ $$\ln y=\ln(\theta+5)-(\ln\theta+\ln\cos\theta)=\ln(\theta+5)-\ln\theta-\ln\cos\theta$$ 2) Then we take the derivative of both sides with respect to $\theta$ and remember that $(\ln \theta)'=1/\theta$: $$\frac{1}{y}\times y'=\frac{1}{\theta+5}(\theta+5)'-\frac{1}{\theta}-\frac{1}{\cos\theta}(\cos\theta)'$$ $$\frac{y'}{y}=\frac{1}{\theta+5}-\frac{1}{\theta}+\frac{\sin\theta}{\cos\theta}$$ $$\frac{y'}{y}=\frac{1}{\theta+5}-\frac{1}{\theta}+\tan\theta$$ $$\frac{y'}{y}=\frac{\theta-(\theta+5)+(\tan\theta)(\theta+5)\theta}{\theta(\theta+5)}$$ $$\frac{y'}{y}=\frac{-5+\theta(\theta+5)\tan\theta}{\theta(\theta+5)}$$ 3) Solve for $y'$: $$y'=\frac{-5+\theta(\theta+5)\tan\theta}{\theta(\theta+5)}\times y$$ 4) Finally, substitute for $y=\frac{\theta+5}{\theta\cos\theta}$ $$y'=\frac{-5+\theta(\theta+5)\tan\theta}{\theta(\theta+5)}\times\frac{\theta+5}{\theta\cos\theta}$$ $$y'=\frac{\theta(\theta+5)\tan\theta-5}{\theta^2\cos\theta}$$ $$y'=\frac{(\theta+5)\tan\theta-5}{\theta\cos\theta}$$