## Calculus: Early Transcendentals (2nd Edition)

$g'(t) = \frac{-3}{t^2} - \frac{2}{t^3}$
First Expand Fraction Into a Sum of Three Fractions $g(t) = \frac{t^3+3t^2+t}{t^3} = 1 + 3t^{-1}+t^{-2}$ Power Rule: $g'(t) = 1(0)t^{(0-1)} + 3(-1)t^{(-1-1)}+1(-2)t^{(-2-1)} = -3t^{-2}-2t^{-3} = \frac{-3}{t^2} - \frac{2}{t^3}$