Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises - Page 149: 37


$f(x)=\sqrt x$ and $a=9$

Work Step by Step

*According to definition, the derivative of a function $f$ at a number $a$ is $$f'(a)=\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Here we have $$\lim\limits_{h\to0}\frac{\sqrt{9+h}-3}{h}$$ $$\lim\limits_{h\to0}\frac{\sqrt{9+h}-\sqrt9}{h}$$ Now we match the formula found above with the formula of the derivative according to definition. We find that $a=9$ and $f(a)=f(9)=\sqrt9$ Therefore, $f(x)=\sqrt{x}$
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