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

$\frac{dr}{d \theta} = 9 (sec \theta - tan \theta)^{\frac{1}{2}} (sec \theta (tan \theta - sec \theta ))$ $\frac{dr}{d \theta} = -9 sec \theta (sec \theta - tan \theta)^{\frac{3}{2}}$
$r =6(u)^{\frac{3}{2}}$ where, $u= (sec \theta - tan \theta)$ Now, $\frac{dr}{du} = 9 u^{\frac{1}{2}}$ and, $\frac{du}{d\theta} = sec \theta \times tan \theta - sec^{2} \theta = sec \theta (tan \theta - sec \theta )$ So, $\frac{dr}{d\ theta} = \frac{dr}{du} \times \frac{du}{d \theta}$ $\frac{dr}{d \theta} = 9 u^{\frac{1}{2}} (sec \theta (tan \theta - sec \theta ))$ $\frac{dr}{d \theta} = 9 (sec \theta - tan \theta)^{\frac{1}{2}} (sec \theta (tan \theta - sec \theta ))$ $\frac{dr}{d \theta} = -9 sec \theta (sec \theta - tan \theta)^{\frac{3}{2}}$