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: 36

Answer

$\displaystyle w' = -\frac{1.4}{z^{2.4}} - \frac{\pi}{2z^{\frac{3}{2}}}$

Work Step by Step

$w =\frac{1}{z^{1.4}} + \frac{\pi}{\sqrt z}$ $w = z^{-1.4} + \pi z^{-\frac{1}{2}}$ $w' = -1.4z^{-2.4} + \pi (-\frac{1}{2}z^{-\frac{3}{2}})$ $w' = -\frac{1.4}{z^{2.4}} - \frac{1}{2} \pi \frac{1}{z^{\frac{3}{2}}}$
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