Answer
$\frac{dr}{d \theta} = 9 (sec \theta - tan \theta)^{\frac{1}{2}} (sec \theta (tan \theta - sec \theta ))$
$\frac{dr}{d \theta} = -9 sec \theta (sec \theta - tan \theta)^{\frac{3}{2}}$
Work Step by Step
$r =6(u)^{\frac{3}{2}}$
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,
$\frac{dr}{d\ theta} = \frac{dr}{du} \times \frac{du}{d \theta}$
$\frac{dr}{d \theta} = 9 u^{\frac{1}{2}} (sec \theta (tan \theta - sec \theta ))$
$\frac{dr}{d \theta} = 9 (sec \theta - tan \theta)^{\frac{1}{2}} (sec \theta (tan \theta - sec \theta ))$
$\frac{dr}{d \theta} = -9 sec \theta (sec \theta - tan \theta)^{\frac{3}{2}}$
