University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises: 12

Answer

$r' = -12x^{-2} +12x^{-4} - 4x^{-5}$ $r'' = 24x^{-3} - 48x^{-5} + 20x^{-6}$

Work Step by Step

$r = \frac{12}{x} - \frac{4}{x^{3}} + \frac{1}{x^{4}}$ (Theta is replaced with x, just for convenience.) $r = 12x^{-1} -4x^{-3} +x^{-4}$ $r' = -12x^{-2} +12x^{-4} - 4x^{-5}$ $r'' = 24x^{-3} - 48x^{-5} + 20x^{-6}$
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