1. **State the problem:** We need to find the derivative of the function $$g(x) = x \cdot 2 + 2 \ln x = 2x + 2 \ln x$$ and then analyze the intervals where the function is increasing or decreasing using a table of changes. 2. **Recall the derivative rules:** - The derivative of $$2x$$ is $$2$$. - The derivative of $$2 \ln x$$ is $$\frac{2}{x}$$ because $$\frac{d}{dx} \ln x = \frac{1}{x}$$. 3. **Calculate the derivative:** $$ g'(x) = 2 + \frac{2}{x} $$ 4. **Find critical points:** Set $$g'(x) = 0$$ to find where the function changes increasing/decreasing behavior: $$ 2 + \frac{2}{x} = 0 \\ 2 = -\frac{2}{x} \\ 2x = -2 \\ x = -1 $$ Since $$x > 0$$ for $$\ln x$$ to be defined, $$x = -1$$ is not in the domain. 5. **Analyze the sign of $$g'(x)$$ for $$x > 0$$:** - For $$x > 0$$, $$\frac{2}{x} > 0$$, so $$g'(x) = 2 + \frac{2}{x} > 0$$. 6. **Conclusion:** - The derivative is positive for all $$x > 0$$, so $$g(x)$$ is increasing on its domain. 7. **Table of changes:** | Interval | Sign of $$g'(x)$$ | Behavior of $$g(x)$$ | |----------|------------------|---------------------| | $$(0, \infty)$$ | Positive | Increasing | **Final answer:** $$g'(x) = 2 + \frac{2}{x}$$ and $$g(x)$$ is increasing for all $$x > 0$$.