Calculus: Early Transcendentals (2nd Edition)

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


$f'(x)=\frac{-1}{x^2}$ $f''(x)=\frac{2}{x^3}$ $f'''(x)=\frac{-6}{x^4}$

Work Step by Step

$f(x)=\frac{1}{x}=x^{-1}$ Using Power Rule: $f'(x) = -x^{-2}=\frac{-1}{x^2}$ $f''(x) = 2x^{-3}=\frac{2}{x^3}$ $f'''(x) = -6x^{-4}=\frac{-6}{x^4}$
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