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

$$y'=\frac{\theta\cos2\theta(1+2\ln\theta)-4\sin\theta\cos\theta}{2\theta(\sin\theta\cos\theta)(1+2\ln\theta)}$$
$$y=\ln\Big(\frac{\sqrt{\sin\theta\cos\theta}}{1+2\ln\theta}\Big)$$ The derivative of $y$: $$y'=\frac{1}{\frac{\sqrt{\sin\theta\cos\theta}}{1+2\ln\theta}}\times\Big(\frac{\sqrt{\sin\theta\cos\theta}}{1+2\ln\theta}\Big)'$$ $$y'=\frac{1+2\ln\theta}{\sqrt{\sin\theta\cos\theta}}\times\frac{(\sqrt{\sin\theta\cos\theta})'(1+2\ln\theta)-(\sqrt{\sin\theta\cos\theta})(1+2\ln\theta)'}{(1+2\ln\theta)^2}$$ We have $$(\sqrt{\sin\theta\cos\theta})'=\frac{1}{2\sqrt{\sin\theta\cos\theta}}(\sin\theta\cos\theta)'=\frac{\cos\theta\cos\theta-\sin\theta\sin\theta}{2\sqrt{\sin\theta\cos\theta}}$$ $$=\frac{\cos^2\theta-\sin^2\theta}{2\sqrt{\sin\theta\cos\theta}}=\frac{\cos2\theta}{2\sqrt{\sin\theta\cos\theta}}$$ and $$(1+2\ln\theta)'=0+\frac{2}{\theta}=\frac{2}{\theta}$$ So, $$y'=\frac{1+2\ln\theta}{\sqrt{\sin\theta\cos\theta}}\times\frac{\frac{\cos2\theta}{2\sqrt{\sin\theta\cos\theta}}(1+2\ln\theta)-\frac{2\sqrt{\sin\theta\cos\theta}}{\theta}}{(1+2\ln\theta)^2}$$ $$y'=\frac{1}{\sqrt{\sin\theta\cos\theta}}\times\frac{\frac{\theta\cos2\theta(1+2\ln\theta)-4\sin\theta\cos\theta}{2\theta\sqrt{\sin\theta\cos\theta}}}{1+2\ln\theta}$$ $$y'=\frac{1}{\sqrt{\sin\theta\cos\theta}}\times\frac{\theta\cos2\theta(1+2\ln\theta)-4\sin\theta\cos\theta}{2\theta\sqrt{\sin\theta\cos\theta}(1+2\ln\theta)}$$ $$y'=\frac{\theta\cos2\theta(1+2\ln\theta)-4\sin\theta\cos\theta}{2\theta(\sin\theta\cos\theta)(1+2\ln\theta)}$$