1. **Problem statement:** Given the function $$g(x) = x - 1 + \ln(x)$$ defined on $$]0, +\infty[,$$ prove that its derivative is $$g'(x) = 1 + \frac{1}{x}$$ for all $$x \in ]0, +\infty[.$$ 2. **Formula and rules:** The derivative of $$g(x)$$ is found by differentiating each term: - The derivative of $$x$$ is 1. - The derivative of $$-1$$ is 0. - The derivative of $$\ln(x)$$ is $$\frac{1}{x}$$ for $$x > 0$$. 3. **Intermediate work:** $$ g'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(1) + \frac{d}{dx}(\ln(x)) = 1 - 0 + \frac{1}{x} = 1 + \frac{1}{x} $$ 4. **Explanation:** The derivative of $$g$$ at any point $$x$$ in its domain is the sum of the derivatives of its components. Since $$x$$ differentiates to 1 and $$\ln(x)$$ differentiates to $$\frac{1}{x}$$, their sum gives the derivative of $$g$$. **Final answer:** $$g'(x) = 1 + \frac{1}{x}$$ for all $$x \in ]0, +\infty[.$$