Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.2 - The Derivative as a Function - Exercises 3.2 - Page 115: 12

Answer

$-\dfrac{w}{(w^2-1)^{3/2}}$

Work Step by Step

Consider $z(w) = \dfrac{1}{\sqrt {(w^{2}-1)}}$ We will use the derivative definition, such as $z'(w) = \lim\limits_{h \to 0}\dfrac{\dfrac{1}{\sqrt {((w+h)^{2}}-1)}-\dfrac{1}{\sqrt {(w^{2}}-1)}}{h}= \lim\limits_{h \to 0}\dfrac{-(h+2w)}{(\sqrt {((w+h)^{2}-1)} \sqrt{(w^{2}-1)})(\sqrt {((w+h)^{2}-1)} +\sqrt {(w^{2}-1))}}$ and $\dfrac{-(0+2w)}{(\sqrt {((w+0)^{2}-1)} \sqrt{(w^{2}-1)})(\sqrt {((w+0)^{2}-1)} +\sqrt {(w^{2}-1))}}=\dfrac{-w}{(w^2-1)\sqrt{w^2-1}}=-\dfrac{w}{(w^2-1)^{3/2}}$
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