Answer
$$r'\left( x \right) = \frac{{ - 8}}{{{{\left( {2x + 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& r\left( x \right) = \frac{{ - 8x}}{{2x + 1}} \cr
& {\text{differentiate both sides}} \cr
& r'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{ - 8x}}{{2x + 1}}} \right] \cr
& {\text{use quotient rule}} \cr
& r'\left( x \right) = \frac{{\left( {2x + 1} \right)\frac{d}{{dx}}\left[ { - 8x} \right] - \left( { - 8x} \right)\left[ {2x + 1} \right]}}{{{{\left( {2x + 1} \right)}^2}}} \cr
& {\text{find derivatives}} \cr
& r'\left( x \right) = \frac{{\left( {2x + 1} \right)\left( { - 8} \right) - \left( { - 8x} \right)\left( x \right)}}{{{{\left( {2x + 1} \right)}^2}}} \cr
& {\text{simplify}} \cr
& r'\left( x \right) = \frac{{ - 16x - 8 + 16x}}{{{{\left( {2x + 1} \right)}^2}}} \cr
& r'\left( x \right) = \frac{{ - 8}}{{{{\left( {2x + 1} \right)}^2}}} \cr} $$
