Answer
$$\frac{{dr}}{{d\theta }} = 3{\theta ^{ - 4}}{\text{ and }}\frac{{{d^2}r}}{{d{\theta ^2}}} = - 12{\theta ^{ - 5}}$$
Work Step by Step
$$\eqalign{
& r = \frac{{\left( {\theta - 1} \right)\left( {{\theta ^2} + \theta + 1} \right)}}{{{\theta ^3}}} \cr
& {\text{Use the distributive property}} \cr
& r = \frac{{\theta \left( {{\theta ^2} + \theta + 1} \right) - \left( {{\theta ^2} + \theta + 1} \right)}}{{{\theta ^3}}} \cr
& r = \frac{{{\theta ^3} + {\theta ^2} + \theta - {\theta ^2} - \theta - 1}}{{{\theta ^3}}} \cr
& r = \frac{{{\theta ^3} - 1}}{{{\theta ^3}}} \cr
& r = \frac{{{\theta ^3}}}{{{\theta ^3}}} - \frac{1}{{{\theta ^3}}} \cr
& {\text{where }}\frac{1}{{{\theta ^n}}} = {\theta ^{ - n}} \cr
& r = 1 - {\theta ^{ - 3}} \cr
& \cr
& {\text{Find the first derivative}} \cr
& \frac{{dr}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {1 - {\theta ^{ - 3}}} \right] \cr
& {\text{use }}\frac{d}{{d\theta }}\left[ {{\theta ^n}} \right] = n{\theta ^{n - 1}} \cr
& \frac{{dr}}{{d\theta }} = 0 - \left( { - 3{\theta ^{ - 4}}} \right) \cr
& \frac{{dr}}{{d\theta }} = 3{\theta ^{ - 4}} \cr
& \cr
& {\text{Find the second derivative}} \cr
& \frac{{{d^2}r}}{{d{\theta ^2}}} = \frac{d}{{d\theta }}\left[ {3{\theta ^{ - 4}}} \right] \cr
& {\text{use }}\frac{d}{{d\theta }}\left[ {{\theta ^n}} \right] = n{\theta ^{n - 1}} \cr
& \frac{{{d^2}r}}{{d{\theta ^2}}} = 3\left( { - 4{\theta ^{ - 5}}} \right) \cr
& \frac{{{d^2}r}}{{d{\theta ^2}}} = - 12{\theta ^{ - 5}} \cr} $$
