Answer
$$\frac{{dr}}{{d\theta }} = - 9\sec \theta {\left( {\sec \theta - \tan \theta } \right)^{3/2}}$$
Work Step by Step
$$\eqalign{
& r = 6{\left( {\sec \theta - \tan \theta } \right)^{3/2}} \cr
& {\text{differentiate with respect to }}\theta \cr
& \frac{{dr}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {6{{\left( {\sec \theta - \tan \theta } \right)}^{3/2}}} \right] \cr
& \cr
& {\text{Using the chain rule }}\frac{d}{{d\theta }}\left[ {{u^n}} \right] = n{u^{n - 1}}\frac{{du}}{{d\theta }} \cr
& \frac{{dr}}{{d\theta }} = 6\left( {\frac{3}{2}} \right){\left( {\sec \theta - \tan \theta } \right)^{3/2 - 1}}\frac{d}{{d\theta }}\left[ {\sec \theta - \tan \theta } \right] \cr
& \frac{{dr}}{{d\theta }} = 9{\left( {\sec \theta - \tan \theta } \right)^{1/2}}\frac{d}{{d\theta }}\left[ {\sec \theta - \tan \theta } \right] \cr
& \cr
& {\text{solving the derivatives}} \cr
& \frac{{dr}}{{d\theta }} = 9{\left( {\sec \theta - \tan \theta } \right)^{1/2}}\left( {\sec \theta \tan \theta - {{\sec }^2}\theta } \right) \cr
& \cr
& {\text{simplifying, we get:}} \cr
& \frac{{dr}}{{d\theta }} = - 9\sec \theta {\left( {\sec \theta - \tan \theta } \right)^{1/2}}\left( {\sec \theta - \tan \theta } \right) \cr
& \frac{{dr}}{{d\theta }} = - 9\sec \theta {\left( {\sec \theta - \tan \theta } \right)^{3/2}} \cr} $$
