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.