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: 48

Answer

$u' = \frac{-3}{x^{4}}$ $u'' = \frac{12}{x^{5}}$

Work Step by Step

$u = \frac{x(x+1)(x^{2}-x+1)}{x^{4}}$ $u = \frac{(x+1)(x^{2}-x+1)}{x^{3}}$ $u = \frac{x^{3}+1}{x^{3}}$ $u' = \frac{x^{3}(3x^{2}) - 3x^{2}(x^{3}+1)}{x^{6}}$ $u' = \frac{-3x^{2}}{x^{6}}$ $u' = \frac{-3}{x^{4}}$ $u'' = \frac{12}{x^{5}}$
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