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: 49

Answer

$w' = \frac{-(z^{2}+1)}{z^{2}}=-1-\frac{1}{z^2}$ $w'' = \frac{2}{z^{3}}$

Work Step by Step

$w = (\frac{1+3z}{3z})(3-z)$ $w = \frac{3+8z-3z^{2}}{3z}$ $w' = \frac{3z(8-6z) - 3(3+8z-3z^{2})}{9z^{2}}$ $w' = \frac{-9(z^{2}+1)}{9z^{2}}$ $w' = \frac{-(z^{2}+1)}{z^{2}}$ $w'' = \frac{-z^{2}(2z)+2z(z^{2}+1)}{z^{4}}$ $w'' = \frac{2z}{z^{4}}$ $w'' = \frac{2}{z^{3}}$
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