Answer
First Derivative: $s'=\frac{2}{t^{2}}-\frac{8}{t^{3}}$
Second Derivative: $s''=-\frac{4}{t^{3}}+\frac{24}{t^{4}}$
Work Step by Step
Using the Power Rule:
First Derivative:
$s=-2t^{-1}+\frac{4}{t^{2}}$
$s'=(-1)(-2)t^{-1-1}+(-2)(4)t^{-2-1}$
$s'=2t^{-2}-8t^{-3}$
$s'=\frac{2}{t^{2}}-\frac{8}{t^{3}}$
Second Derivative:
$s'=2t^{-2}-8t^{-3}$
$s''=(-2)(2)t^{-2-1}-(-3)(8)t^{-3-1}$
$s''=-4t^{-3}+24t^{-4}$
$s''=-\frac{4}{t^{3}}+\frac{24}{t^{4}}$
