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We are tasked with finding the derivative \( f'(x) \) of the function \( f(x) = \sqrt{2x + 4} \) using the limit definition of the derivative:

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

### Step-by-Step Solution:

1. **Substitute \( f(x+h) \) and \( f(x) \):**
   Given \( f(x) = \sqrt{2x + 4} \), we first calculate \( f(x+h) \):
   \[
   f(x+h) = \sqrt{2(x+h) + 4} = \sqrt{2x + 2h + 4}.
   \]

2. **Set up the limit definition of the derivative:**
   \[
   f'(x) = \lim_{h \to 0} \frac{\sqrt{2x + 2h + 4} - \sqrt{2x + 4}}{h}.
   \]

3. **Rationalize the numerator:**
   To simplify this expression, multiply both the numerator and denominator by the conjugate of the numerator:
   \[
   f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{2x + 2h + 4} - \sqrt{2x + 4}\right)}{h} \times \frac{\left(\sqrt{2x + 2h + 4} + \sqrt{2x + 4}\right)}{\left(\sqrt{2x + 2h + 4} + \sqrt{2x + 4}\right)}.
   \]
   This simplifies the numerator using the difference of squares:
   \[
   f'(x) = \lim_{h \to 0} \frac{(2x + 2h + 4) - (2x + 4)}{h \left(\sqrt{2x + 2h + 4} + \sqrt{2x + 4}\right)}.
   \]
   Simplifying further:
   \[
   f'(x) = \lim_{h \to 0} \frac{2h}{h \left(\sqrt{2x + 2h + 4} + \sqrt{2x + 4}\right)}.
   \]

4. **Cancel out \( h \) from the numerator and denominator:**
   \[
   f'(x) = \lim_{h \to 0} \frac{2}{\sqrt{2x + 2h + 4} + \sqrt{2x + 4}}.
   \]

5. **Evaluate the limit as \( h \to 0 \):**
   As \( h \) approaches 0, \( 2x + 2h + 4 \) approaches \( 2x + 4 \). Therefore, we can simplify the expression:
   \[
   f'(x) = \frac{2}{\sqrt{2x + 4} + \sqrt{2x + 4}} = \frac{2}{2\sqrt{2x + 4}}.
   \]

6. **Final simplification:**
   \[
   f'(x) = \frac{1}{\sqrt{2x + 4}}.
   \]

### Conclusion:
The derivative of \( f(x) = \sqrt{2x + 4} \) is:

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

Would you like any further clarifications or details?

Here are some related questions to explore:

1. How would the derivative change if the function was \( f(x) = \sqrt{3x + 5} \)?
2. Can you apply the same process to find the derivative of \( f(x) = \sqrt{x^2 + 4} \)?
3. What is the significance of the conjugate in rationalizing the numerator in this problem?
4. How does this derivative compare with the derivative of a linear function?
5. What would the second derivative of \( f(x) = \sqrt{2x + 4} \) be?

**Tip**: Rationalizing the numerator is a common technique when dealing with square roots in limit problems.

Let f(x) = sqrt(2x + 4). Find f'(x) by evaluating lim(h->0) [f(x+h) - f(x)] / h.

This problem involves finding the derivative of f(x) = sqrt(2x + 4) by applying the limit definition. The solution involves rationalizing the numerator, applying limits, and simplifying square roots. Suitable for advanced high school students learning calculus.

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