Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.1 Introducing the Derivative - 3.1 Execises: 43

Answer

The value is $$\left.\frac{dy}{dt}\right|_{t=1}=-\frac{1}{4}.$$

Work Step by Step

Using the definition of the derivative and putting $t=1$ we have $$\left.\frac{dy}{dt}\right|_{t=1}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{\frac{1}{(1+h)+1}-\frac{1}{1+1}}{h}=\lim_{h\to0}\frac{\frac{1}{2+h}-\frac{1}{2}}{h}=\lim_{h\to0}\frac{\frac{2-(2+h)}{2(2+h)}}{h}=\lim_{h\to0}\frac{-h}{2h(2+h)}=\lim_{h\to0}\frac{-1}{2(2+h)}=\frac{-1}{2(2+0)}=\frac{-1}{4}.$$
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