1. **State the problem:** We want to find the derivative of the function $f(x) = \sqrt{x+2}$ from first principles. 2. **Recall the definition of derivative from first principles:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Apply the definition to our function:** $$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h+2} - \sqrt{x+2}}{h}$$ 4. **Rationalize the numerator to simplify:** Multiply numerator and denominator by the conjugate: $$\frac{\sqrt{x+h+2} - \sqrt{x+2}}{h} \times \frac{\sqrt{x+h+2} + \sqrt{x+2}}{\sqrt{x+h+2} + \sqrt{x+2}} = \frac{(x+h+2) - (x+2)}{h(\sqrt{x+h+2} + \sqrt{x+2})}$$ 5. **Simplify the numerator:** $$(x+h+2) - (x+2) = h$$ 6. **Substitute back:** $$\frac{h}{h(\sqrt{x+h+2} + \sqrt{x+2})} = \frac{1}{\sqrt{x+h+2} + \sqrt{x+2}}$$ 7. **Take the limit as $h \to 0$:** $$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h+2} + \sqrt{x+2}} = \frac{1}{2\sqrt{x+2}}$$ 8. **Final answer:** $$\boxed{f'(x) = \frac{1}{2\sqrt{x+2}}}$$ This shows from first principles that the derivative of $\sqrt{x+2}$ is $\frac{1}{2\sqrt{x+2}}$.