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 $$y'=(\ln3x+x)'=\frac{1}{3x}(3x)'+1=\frac{3}{3x}+1$$ $$y'=\frac{1}{x}+1$$
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