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: 3

Answer

$g'(-1) = 2$ $g'(2) = -\frac{1}{4}$ $g'(\sqrt 3) = -\frac{2}{3\sqrt 3}$

Work Step by Step

$g(t) = \frac{1}{t^{2}}$ Using definition of derivative: $g'(t) = \lim\limits_{h \to 0}\frac{g(t+h) - g(t)}{h}$ $g'(t) = \lim\limits_{h \to 0}\frac{\frac{1}{(t+h)^{2} - }\frac{1}{t^{2}}}{h}$ $g'(t) = \lim\limits_{h \to 0}\frac{-(2t+h)}{t^{4}+2t^{3}h+h^{2}t^{2}}$ $g'(t) = -2\frac{1}{t^{3}}$ Now, $g'(-1) = 2$ $g'(2) = -\frac{1}{4}$ $g'(\sqrt 3) = -\frac{2}{3\sqrt 3}$
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