Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 41

Answer

$g'(t) = \frac{-3}{t^2} - \frac{2}{t^3}$

Work Step by Step

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: $g'(t) = 1(0)t^{(0-1)} + 3(-1)t^{(-1-1)}+1(-2)t^{(-2-1)} = -3t^{-2}-2t^{-3} = \frac{-3}{t^2} - \frac{2}{t^3}$
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