Answer
$$y'=\frac{1}{x}+1$$
Work Step by Step
$$y=\ln3x+x$$
Recall the Derivative Rule for Natural Logarithm: $$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$
We have $$y'=(\ln3x+x)'=\frac{1}{3x}(3x)'+1=\frac{3}{3x}+1$$ $$y'=\frac{1}{x}+1$$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 174: 11
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