Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.8 - The Derivative as a Function - 2.8 Exercises - Page 162: 34

Answer

(a) $f'(x) = 1-\frac{1}{x^2}$ (b) When we compare the graphs of $f(x) = x+\frac{1}{x}$ and $f'(x) = 1-\frac{1}{x^2}$, the answer to part (a) seems reasonable.

Work Step by Step

(a) $f(x) = x + \frac{1}{x}$ We can find $f'(x)$: $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$ $f'(x) = \lim\limits_{h \to 0} \frac{(x+h)+\frac{1}{x+h}-(x+\frac{1}{x})}{h}$ $f'(x) = \lim\limits_{h \to 0} \frac{h+\frac{1}{x+h}-\frac{1}{x}}{h}$ $f'(x) = \lim\limits_{h \to 0} \frac{\frac{hx(x+h)+x-(x+h)}{(x)(x+h)}}{h}$ $f'(x) = \lim\limits_{h \to 0} \frac{hx(x+h)-h}{(h)(x)(x+h)}$ $f'(x) = \lim\limits_{h \to 0} \frac{x(x+h)-1}{(x)(x+h)}$ $f'(x) = \frac{x(x+0)-1}{(x)(x+0)}$ $f'(x) = \frac{x^2-1}{x^2}$ $f'(x) = 1-\frac{1}{x^2}$ (b) When we compare the graphs of $f(x) = x+\frac{1}{x}$ and $f'(x) = 1-\frac{1}{x^2}$, the answer to part (a) seems reasonable.
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