  Use the x − a definition of a derivative (p. 144, Equation 5) to find f
0
(2) when f(x) = √
x

To find the derivative of \( f(x) = \sqrt{x} \) at \( x = 2 \) using the definition of the derivative based on the difference \( x - a \) (also known as the "difference quotient" definition), we follow this procedure:

The derivative \( f'(a) \) is given by the formula:

\[
f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}
\]

Here, \( f(x) = \sqrt{x} \) and \( a = 2 \). So, we want to compute \( f'(2) \). First, apply the definition of the derivative:

\[
f'(2) = \lim_{x \to 2} \frac{\sqrt{x} - \sqrt{2}}{x - 2}
\]

### Step 1: Simplify the expression
The direct substitution will lead to an indeterminate form \( \frac{0}{0} \), so we must simplify the expression by multiplying both the numerator and denominator by the conjugate of the numerator:

\[
f'(2) = \lim_{x \to 2} \frac{\sqrt{x} - \sqrt{2}}{x - 2} \cdot \frac{\sqrt{x} + \sqrt{2}}{\sqrt{x} + \sqrt{2}}
\]

### Step 2: Multiply and simplify
Now, simplify the numerator using the difference of squares formula:

\[
f'(2) = \lim_{x \to 2} \frac{(\sqrt{x})^2 - (\sqrt{2})^2}{(x - 2)(\sqrt{x} + \sqrt{2})}
\]

Since \( (\sqrt{x})^2 = x \) and \( (\sqrt{2})^2 = 2 \), we get:

\[
f'(2) = \lim_{x \to 2} \frac{x - 2}{(x - 2)(\sqrt{x} + \sqrt{2})}
\]

### Step 3: Cancel common factors
Cancel the factor \( x - 2 \) in both the numerator and denominator:

\[
f'(2) = \lim_{x \to 2} \frac{1}{\sqrt{x} + \sqrt{2}}
\]

### Step 4: Evaluate the limit
Now substitute \( x = 2 \) into the remaining expression:

\[
f'(2) = \frac{1}{\sqrt{2} + \sqrt{2}} = \frac{1}{2\sqrt{2}}
\]

### Step 5: Simplify the result
Simplify the fraction:

\[
f'(2) = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}
\]

Thus, the derivative of \( f(x) = \sqrt{x} \) at \( x = 2 \) is:

\[
f'(2) = \frac{\sqrt{2}}{4}
\]

Would you like further clarification on any of the steps?

### Follow-up Questions:
1. How can the limit definition of a derivative be applied to more complex functions?
2. What is the geometric interpretation of the derivative at a point?
3. Can we use this method to find the derivative of any radical function?
4. How does this method compare to using standard differentiation rules?
5. What would happen if we attempted this process at a point where \( f(x) \) is not differentiable?

### Tip:
Multiplying by the conjugate is a common technique to simplify limits involving square roots. It often helps remove indeterminate forms.

Use the x − a definition of a derivative (p. 144, Equation 5) to find f'(2) when f(x) = √x.

Learn how to find the derivative of f(x) = √x at x = 2 using the x − a definition of a derivative. This step-by-step guide explains key calculus concepts such as limits and the difference of squares, providing a clear understanding for advanced high school students.

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