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: 9

Answer

$s'(t) = \frac{1}{(2t+1)^{2}}$

Work Step by Step

$s(t) = \frac{t}{2t+1}$ Using the derivative definition: $s'(t) = \lim\limits_{h \to 0}\frac{\frac{t+h}{2(t+h)+1} - \frac{t}{2t+1}}{h}$ $s'(t) = \lim\limits_{h \to 0}\frac{h}{h(2t+1)^{2}}$ $s'(t) = \frac{1}{(2t+1)^{2}}$
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