To find $f'(x)$ for $f(x) = \sqrt{x}$ using the definition of the derivative, we proceed as follows:

Step-by-Step Solution

The definition of the derivative is given by: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

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

2. Rationalize the numerator: Multiply both the numerator and the 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 expression: Using the difference of squares, $(a - b)(a + b) = a^2 - b^2$, we get: $f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$ Simplify further: $f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$

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 $h \to 0$: Substitute $h = 0$: $f'(x) = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$

Final Answer

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

Would you like a more detailed explanation, or do you have any questions?

Here are five related questions to explore this topic further:

1. How would you find the derivative of $f(x) = x^{1/3}$ using the definition of the derivative?
2. Can you verify $f'(x) = \frac{1}{2\sqrt{x}}$ by using the power rule instead?
3. How can you use the chain rule to find the derivative of a composite function involving square roots, like $f(x) = \sqrt{g(x)}$?
4. What would be the derivative of $f(x) = \sqrt{x+2}$ using the definition of the derivative?
5. How does the derivative of $f(x) = \sqrt{x}$ behave as $x$ approaches zero?

Tip: Rationalizing the numerator is a useful technique when dealing with square roots in limits, especially for finding derivatives of root functions.