University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises: 12

Answer

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

Work Step by Step

$z(w) = \frac{1}{\sqrt (w^{2}-1)}$ Using the derivative definition: $z'(w) = \lim\limits_{h \to 0}\frac{\frac{1}{\sqrt ((w+h)^{2}-1)}-\frac{1}{\sqrt (w^{2}-1)}}{h}$ $z(w) = \lim\limits_{h \to 0}\frac{(w^{2}-1) - ((w+h)^{2}-1)}{h(\sqrt((w+h)^{2}-1) \sqrt(w^{2}-1))(\sqrt ((w+h)^{2}-1) +\sqrt (w^{2}-1))}$ $z(w) = \lim\limits_{h \to 0}\frac{-(h+2w)}{(\sqrt((w+h)^{2}-1) \sqrt(w^{2}-1))(\sqrt ((w+h)^{2}-1) +\sqrt (w^{2}-1))}$ $z(w) = \frac{-(2w)}{2(w^{2}-1)^{\frac{3}{2}}}$ $z(w) = \frac{-w}{(w^{2}-1)^{\frac{3}{2}}}$
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