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: 9



Work Step by Step

Consider $s(t) = \dfrac{t}{2t+1}$ We will use the derivative definition, such that $s'(t) = \lim\limits_{h \to 0}\dfrac{\dfrac{(t+h)}{2(t+h)+1} - \dfrac{t}{(2t+1)}}{h}= \lim\limits_{h \to 0}[\dfrac{h}{h(2t+1)^{2}}]=\dfrac{1}{(2t+1)^{2}}$
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