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 - Page 133: 45


The value of the derivative is $$\left.\frac{dc}{dt}\right|_{s=25}=\frac{1}{5}.$$

Work Step by Step

Using the definition of the derivative for $s=25$ we have $$\left.\frac{dc}{dt}\right|_{s=25}=\lim_{h\to0}\frac{c(25+h)-c(25)}{h}=\lim_{h\to0}\frac{2\sqrt{25+h}-1-(2\sqrt{25}-1)}{h}=\lim_{h\to0}\frac{2\sqrt{25+h}-10}{h}=\lim_{h\to0}\frac{2\sqrt{25+h}-10}{h}\cdot\frac{2\sqrt{25+h}+10}{2\sqrt{25+h}+10}=\lim_{h\to0}\frac{(2\sqrt{25+h})^2-10^2}{h(2\sqrt{25+h}+10)}=\lim_{h\to0}\frac{4(25+h)-100}{h(2\sqrt{25+h}+10)}=\lim_{h\to0}\frac{4h}{h(2\sqrt{25+h}+10)}=\lim_{h\to0}\frac{4}{2\sqrt{25+0}+10}=\frac{1}{5}.$$
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