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: 18

Answer

$z' = \frac{9x^{2}-24x-4}{(3x^{2}+x)^{2}}$

Work Step by Step

$z = \frac{4-3x}{3x^{2}+x}$ $z' = \frac{(3x^{2}+x)(-3) - (4-3x)(6x+1)}{(3x^{2}+x)^{2}}$ $z'= \frac{-9x^{2}-3x-24x-4+18x^{2}+3x}{(3x^{2}+x)^{2}}$ $z' = \frac{9x^{2}-24x-4}{(3x^{2}+x)^{2}}$
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