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

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