Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 223: 5

Answer

$f'(x) = -\frac{1}{x}$

Work Step by Step

$f(x) = \ln \frac{1}{x}$ $f(x) = \ln (x^{-1})$ Formula: $\ln x^{a} = a \ln x$ $f(x) = -\ln x$ Differentiate: $f'(x) = \frac{d(-\ln x)}{dx}$ Formula: $\frac{d}{dx} (\ln x) = \frac{1}{x}$ $f'(x) = -\frac{1}{x}$
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