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: 31

Answer

= 2

Work Step by Step

We use the rule for taking the derivative of a natural log and chain rule to obtain: f(x) = ln(x + lnx) f'(x) = $\frac{(x + lnx)'}{x + lnx}$ f'(x) = $\frac{1 + \frac{1}{x}}{x + lnx}$ f'(1) = $\frac{1 + \frac{1}{1}}{1 + ln1}$ f'(1) = $\frac{1 +1}{1 + 0}$ f'(1) = $\frac{2}{1}$ f'(1) = 2
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