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: 35

Answer

$f(x) = sin~x+ln~x$ $f'(x) = cos~x+\frac{1}{x}$ We can see a sketch of the graphs below.
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Work Step by Step

$f(x) = sin~x+ln~x$ $f'(x) = cos~x+\frac{1}{x}$ $f'(x)$ seems reasonable since $f'(x)$ has negative values when the slope of $f(x)$ has a negative slope, $f'(x)$ is 0 when the slope of $f(x)$ is 0, and $f'(x)$ has positive values when the slope of $f(x)$ has a positive slope.
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