1. **State the problem:** We want to find the derivative of the function $y = x + 3$ using the first principle of derivatives. 2. **Recall the formula for the derivative using first principles:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Apply the function to the formula:** Given $f(x) = x + 3$, then $$f(x+h) = (x+h) + 3 = x + h + 3$$ 4. **Substitute into the difference quotient:** $$\frac{f(x+h) - f(x)}{h} = \frac{(x + h + 3) - (x + 3)}{h}$$ 5. **Simplify the numerator:** $$\frac{\cancel{x} + h + 3 - \cancel{x} - 3}{h} = \frac{h}{h}$$ 6. **Simplify the fraction:** $$\frac{\cancel{h}}{\cancel{h}} = 1$$ 7. **Take the limit as $h$ approaches 0:** $$f'(x) = \lim_{h \to 0} 1 = 1$$ **Final answer:** The derivative of $y = x + 3$ is $$\boxed{1}$$