To differentiate $y = \sqrt{x}$ using first principles, we use the definition of the derivative:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Given $f(x) = \sqrt{x}$, let's substitute this into the formula:

1. Substitute $f(x) = \sqrt{x}$:

$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$

2. Multiply the numerator and denominator by the conjugate of the numerator $\sqrt{x+h} + \sqrt{x}$:

$f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}$

3. Simplify the numerator using the difference of squares:

$(\sqrt{x+h})^2 - (\sqrt{x})^2 = (x+h) - x = h$

Now, the expression becomes:

$f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$

4. Cancel $h$ in the numerator and denominator:

$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$

5. Take the limit as $h \to 0$:

As $h \to 0$, $\sqrt{x+h} \to \sqrt{x}$, so:

$f'(x) = \frac{1}{2\sqrt{x}}$

Thus, the derivative of $y = \sqrt{x}$ using first principles is:

$\boxed{f'(x) = \frac{1}{2\sqrt{x}}}$

Would you like more details on this?

Here are 5 related questions:

1. How would you differentiate $y = \frac{1}{x}$ using first principles?
2. How can you apply first principles to differentiate $y = x^n$?
3. What are the advantages of using the conjugate when simplifying expressions with square roots?
4. How do you differentiate $y = \ln(x)$ using first principles?
5. Can you explain the concept of limits in the definition of a derivative?

Tip: The first principles method gives you an intuitive understanding of the derivative as the slope of the tangent line.