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: 25

Answer

$\frac{-4}{(2t-2)^2}$

Work Step by Step

$\left(\frac{f}{g}\right)'$ = $\frac{f'g-fg'}{g^2}$ $(3t-1)(2t-2)^{-1}$ = $\frac{3t-1}{2t-2}$ $(\frac{3t-1}{2t-2})$ = $\frac{(3)(2t-2)-(3t-1)(2)}{(2t-2)^2}$ = $\frac{-4}{(2t-2)^2}$
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