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

$$\frac{dr}{d\theta}=e^\theta\Big(\frac{\theta-2}{\theta^3}+\frac{2\theta-\pi}{2\theta^{(\pi/2)+1}}\Big)$$
$$r=e^{\theta}\Big(\frac{1}{\theta^2}+\theta^{-\pi/2}\Big)=e^\theta\Big(\theta^{-2}+\theta^{-\pi/2}\Big)$$ Apply the Derivative Product Rule here, we have $$\frac{dr}{d\theta}=(e^\theta)'\Big(\theta^{-2}+\theta^{-\pi/2}\Big)+e^\theta\Big(\theta^{-2}+\theta^{-\pi/2}\Big)'$$ $$\frac{dr}{d\theta}=e^\theta\Big(\theta^{-2}+\theta^{-\pi/2}\Big)+e^\theta\Big((\theta^{-2})'+(\theta^{-\pi/2})'\Big)$$ $$\frac{dr}{d\theta}=e^\theta\Big(\theta^{-2}+\theta^{-\pi/2}\Big)+e^\theta\Big(-2\theta^{-3}-\frac{\pi}{2}\theta^{(-\pi/2)-1}\Big)$$ $$\frac{dr}{d\theta}=e^\theta\Big(\theta^{-2}+\theta^{-\pi/2}\Big)+e^\theta\Big(-2\theta^{-3}-\frac{\pi}{2}\theta^{(-\pi-2)/2}\Big)$$ $$\frac{dr}{d\theta}=e^{\theta}\Big(\theta^{-2}+\theta^{-\pi/2}-2\theta^{-3}-\frac{\pi}{2}\theta^{(-\pi-2)/2}\Big)$$ $$\frac{dr}{d\theta}=e^\theta\Big(\frac{1}{\theta^2}-\frac{2}{\theta^3}+\frac{1}{\theta^{\pi/2}}-\frac{\pi}{2\theta^{(\pi+2)/2}}\Big)$$ $$\frac{dr}{d\theta}=e^\theta\Big(\frac{1}{\theta^2}-\frac{2}{\theta^3}+\frac{1}{\theta^{\pi/2}}-\frac{\pi}{2\theta^{(\pi/2)+1}}\Big)$$ $$\frac{dr}{d\theta}=e^\theta\Big(\frac{\theta-2}{\theta^3}+\frac{2\theta-\pi}{2\theta^{(\pi/2)+1}}\Big)$$