1. **State the problem:** Find the derivative of the function $$f(x) = \sqrt{x} - 1 - \sqrt{x} \ln(x)$$. 2. **Recall the formulas and rules:** - Derivative of $$\sqrt{x} = x^{1/2}$$ is $$\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$$. - Derivative of $$\ln(x)$$ is $$\frac{1}{x}$$. - Use the product rule for $$\sqrt{x} \ln(x)$$: if $$u = \sqrt{x}$$ and $$v = \ln(x)$$, then $$\frac{d}{dx}(uv) = u'v + uv'$$. 3. **Calculate derivatives:** - Derivative of $$\sqrt{x} - 1$$ is $$\frac{1}{2\sqrt{x}} - 0 = \frac{1}{2\sqrt{x}}$$. - For $$\sqrt{x} \ln(x)$$: - $$u = x^{1/2}$$, so $$u' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$$. - $$v = \ln(x)$$, so $$v' = \frac{1}{x}$$. - Apply product rule: $$\frac{d}{dx}(\sqrt{x} \ln(x)) = \frac{1}{2\sqrt{x}} \ln(x) + \sqrt{x} \cdot \frac{1}{x}$$. 4. **Write the derivative of the whole function:** $$f'(x) = \frac{1}{2\sqrt{x}} - \left( \frac{1}{2\sqrt{x}} \ln(x) + \frac{\sqrt{x}}{x} \right)$$. 5. **Simplify:** Note that $$\frac{\sqrt{x}}{x} = \frac{x^{1/2}}{x^{1}} = x^{-1/2} = \frac{1}{\sqrt{x}}$$. So, $$f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{x}} \ln(x) - \frac{1}{\sqrt{x}}$$. 6. **Combine like terms:** $$f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{\sqrt{x}} - \frac{1}{2\sqrt{x}} \ln(x) = -\frac{1}{2\sqrt{x}} \ln(x) + \left( \frac{1}{2\sqrt{x}} - \frac{1}{\sqrt{x}} \right)$$. Simplify the constants: $$\frac{1}{2\sqrt{x}} - \frac{1}{\sqrt{x}} = -\frac{1}{2\sqrt{x}}$$. 7. **Final derivative:** $$f'(x) = -\frac{1}{2\sqrt{x}} \ln(x) - \frac{1}{2\sqrt{x}} = -\frac{1}{2\sqrt{x}} (\ln(x) + 1)$$. **Answer:** $$\boxed{f'(x) = -\frac{\ln(x) + 1}{2\sqrt{x}}}$$