1. **State the problem:** Find the derivative of the function $f(x) = \sqrt{x} + 1$ using the first principle of derivatives (definition of derivative). 2. **Recall the definition of derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Apply the definition to $f(x)$:** $$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} + 1 - (\sqrt{x} + 1)}{h} = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$ 4. **Simplify the numerator by rationalizing:** Multiply numerator and denominator by the conjugate $\sqrt{x+h} + \sqrt{x}$: $$f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$$ 5. **Simplify the fraction:** $$f'(x) = \lim_{h \to 0} \frac{\cancel{h}}{\cancel{h}(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$$ 6. **Evaluate the limit as $h \to 0$:** $$f'(x) = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$$ 7. **Final answer:** $$\boxed{f'(x) = \frac{1}{2\sqrt{x}}}$$ This means the derivative of $f(x) = \sqrt{x} + 1$ is $\frac{1}{2\sqrt{x}}$. The constant 1 disappears because the derivative of a constant is zero.