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: 19

Answer

$g' = \frac{x^{2}+x+4}{(x+0.5)^{2}}$

Work Step by Step

$g = \frac{x^{2}-4}{x+0.5}$ $g' = \frac{(x+0.5)(2x)- (x^{2}-4)(1)}{(x+0.5)^{2}}$ $g' = \frac{x^{2}+x+4}{(x+0.5)^{2}}$
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