Answer
$f'(x)=\frac{-1}{x^2}$
$f''(x)=\frac{2}{x^3}$
$f'''(x)=\frac{-6}{x^4}$
Work Step by Step
$f(x)=\frac{1}{x}=x^{-1}$
Using Power Rule:
$f'(x) = -x^{-2}=\frac{-1}{x^2}$
$f''(x) = 2x^{-3}=\frac{2}{x^3}$
$f'''(x) = -6x^{-4}=\frac{-6}{x^4}$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises - Page 161: 62
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