Answer
$r'\left( t \right) = - \frac{{2a}}{{{t^3}}} - \frac{{4b}}{{{t^5}}}$
Work Step by Step
$$\eqalign{
& r\left( t \right) = \frac{a}{{{t^2}}} + \frac{b}{{{t^4}}} \cr
& {\text{Rewrite the function}}{\text{, apply the property }}\frac{1}{{{t^n}}} = {t^{ - n}} \cr
& r\left( t \right) = a{t^{ - 2}} + b{t^{ - 4}} \cr
& {\text{Differentiate the function}} \cr
& r'\left( t \right) = \frac{d}{{dt}}\left[ {a{t^{ - 2}} + b{t^{ - 4}}} \right] \cr
& {\text{Use the sum diffence rules for differentiation }} \cr
& r'\left( t \right) = \frac{d}{{dt}}\left[ {a{t^{ - 2}}} \right] + \frac{d}{{dt}}\left[ {b{t^{ - 4}}} \right] \cr
& {\text{Use the constant multiple rule}} \cr
& r'\left( t \right) = a\frac{d}{{dt}}\left[ {{t^{ - 2}}} \right] + b\frac{d}{{dt}}\left[ {{t^{ - 4}}} \right] \cr
& {\text{Apply the power rule: }}\frac{d}{{dt}}\left[ {{t^n}} \right] = n{t^{n - 1}} \cr
& r'\left( t \right) = a\left( { - 2{t^{ - 2 - 1}}} \right) + b\left( { - 4{t^{ - 4 - 1}}} \right) \cr
& {\text{Simplify}} \cr
& r'\left( t \right) = - 2a{t^{ - 3}} - 4b{t^{ - 5}} \cr
& {\text{Rewrite}} \cr
& r'\left( t \right) = - \frac{{2a}}{{{t^3}}} - \frac{{4b}}{{{t^5}}} \cr} $$
