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