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: 4


$\displaystyle k'(z) = \frac{-1}{2z^{2}}$ $\displaystyle k'(-1) = -\frac{1}{2}$ $\displaystyle k'(1) = -\frac{1}{2}$ $\displaystyle k'(\sqrt 2) = -\frac{1}{4}$

Work Step by Step

$k(z) = \frac{1-z}{2z}$ Using the derivative definition: $k'(z) = \lim\limits_{h \to 0}\frac{\frac{1-(z+h)}{2(z+h)} - \frac{1-z}{2z}}{h}$ $k'(z) = \lim\limits_{h \to 0}\frac{1}{h}\frac{-2h}{4z(z+h)}$ $k'(z) = \lim\limits_{h \to 0}\frac{-1}{2z(z+h)}$ $k'(z) = \frac{-1}{2z^{2}}$ Now, $k'(-1) = -\frac{1}{2}$ $k'(1) = -\frac{1}{2}$ $k'(\sqrt 2) = -\frac{1}{4}$
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