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: 34

Answer

$$f'(a)=-2a^{-3}$$

Work Step by Step

$$f(x)=x^{-2}$$ That means $$f(a)=a^{-2}$$ The derivative of $f(x)$ at number $a$ can be found as follows: $$f'(a)=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{x^{-2}-a^{-2}}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{\frac{1}{x^2}-\frac{1}{a^2}}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{\frac{a^2-x^2}{a^2x^2}}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{-(x^2-a^2)}{a^2x^2(x-a)}$$ $$f'(a)=-\lim\limits_{x\to a}\frac{(x-a)(x+a)}{a^2x^2(x-a)}$$ $$f'(a)=-\lim\limits_{x\to a}\frac{x+a}{a^2x^2}$$ $$f'(a)=-\frac{a+a}{a^2a^2}$$ $$f'(a)=-\frac{2a}{a^4}=-\frac{2}{a^3}=-2a^{-3}$$
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