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: 35



Work Step by Step

$$f(x)=\sqrt{1-2x}$$ That means $$f(a)=\sqrt{1-2a}$$ The derivative of $f(x)$ at number $a$ can be found as follows: $$f'(a)=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{\sqrt{1-2x}-\sqrt{1-2a}}{x-a}$$ Multiply both numerator and denominator by $\sqrt{1-2x}+\sqrt{1-2a}$, the numerator would become $$(\sqrt{1-2x}-\sqrt{1-2a})(\sqrt{1-2x}+\sqrt{1-2a})$$$$=(1-2x)-(1-2a)$$$$=2a-2x$$$$=-2(x-a)$$ Therefore, $$f'(a)=\lim\limits_{x\to a}\frac{-2(x-a)}{(x-a)(\sqrt{1-2x}+\sqrt{1-2a})}$$ $$f'(a)=\lim\limits_{x\to a}\frac{-2}{\sqrt{1-2x}+\sqrt{1-2a}}$$ $$f'(a)=\frac{-2}{\sqrt{1-2a}+\sqrt{1-2a}}$$ $$f'(a)=\frac{-2}{2\sqrt{1-2a}}$$ $$f'(a)=\frac{-1}{\sqrt{1-2a}}$$
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