1. **State the problem:** Differentiate the function $y=\sqrt{2x+1}$ using first principles (definition of derivative). 2. **Recall the definition of derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ This means we find the slope of the secant line between points $x$ and $x+h$ and then take the limit as $h$ approaches zero. 3. **Apply the definition to $y=\sqrt{2x+1}$:** $$f'(x) = \lim_{h \to 0} \frac{\sqrt{2(x+h)+1} - \sqrt{2x+1}}{h} = \lim_{h \to 0} \frac{\sqrt{2x+2h+1} - \sqrt{2x+1}}{h}$$ 4. **Rationalize the numerator to simplify:** Multiply numerator and denominator by the conjugate: $$\frac{\sqrt{2x+2h+1} - \sqrt{2x+1}}{h} \times \frac{\sqrt{2x+2h+1} + \sqrt{2x+1}}{\sqrt{2x+2h+1} + \sqrt{2x+1}}$$ 5. **Simplify numerator using difference of squares:** $$= \frac{(2x+2h+1) - (2x+1)}{h \left(\sqrt{2x+2h+1} + \sqrt{2x+1}\right)} = \frac{2h}{h \left(\sqrt{2x+2h+1} + \sqrt{2x+1}\right)}$$ 6. **Cancel $h$ in numerator and denominator:** $$= \frac{\cancel{2h}}{\cancel{h} \left(\sqrt{2x+2h+1} + \sqrt{2x+1}\right)} = \frac{2}{\sqrt{2x+2h+1} + \sqrt{2x+1}}$$ 7. **Take the limit as $h \to 0$:** $$f'(x) = \lim_{h \to 0} \frac{2}{\sqrt{2x+2h+1} + \sqrt{2x+1}} = \frac{2}{\sqrt{2x+1} + \sqrt{2x+1}} = \frac{2}{2\sqrt{2x+1}} = \frac{1}{\sqrt{2x+1}}$$ 8. **Final answer:** $$\boxed{f'(x) = \frac{1}{\sqrt{2x+1}}}$$ This means the derivative of $y=\sqrt{2x+1}$ using first principles is $\frac{1}{\sqrt{2x+1}}$.