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: 15

Answer

$f'(x) = 6x+1$

Work Step by Step

Using Product Rule: $f(x) = (x-1)(3x+4)$ $f'(x) = (1)(3x+4)+(x-1)(3) = 3x+4+3x-3=6x+1$ Expansion First and then Power Rule: $f(x) = (x-1)(3x+4) = 3x^2+x-4$ $f'(x)=3(2)x^{2-1}+(1)x^{1-1}-4(0)x^{0-1} = 6x+1$
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