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To solve this problem, we need to use the **limit definition of the derivative** to find the derivative of the function \( f(x) = \sqrt{2x + 1} \) at \( x = 4 \).

The limit definition of the derivative is:

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

In this case, we are asked to find the derivative at \( x = 4 \), so we will apply the definition to:

\[
f'(4) = \lim_{h \to 0} \frac{f(4+h) - f(4)}{h}
\]

### Step-by-Step Solution:

1. **Find \( f(4) \):**
   \[
   f(4) = \sqrt{2(4) + 1} = \sqrt{8 + 1} = \sqrt{9} = 3
   \]

2. **Find \( f(4+h) \):**
   \[
   f(4+h) = \sqrt{2(4+h) + 1} = \sqrt{8 + 2h + 1} = \sqrt{9 + 2h}
   \]

3. **Set up the difference quotient:**
   \[
   f'(4) = \lim_{h \to 0} \frac{\sqrt{9 + 2h} - 3}{h}
   \]

4. **Simplify using conjugates:**
   To simplify the expression, multiply both the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{9 + 2h} + 3 \):

   \[
   f'(4) = \lim_{h \to 0} \frac{\left( \sqrt{9 + 2h} - 3 \right) \left( \sqrt{9 + 2h} + 3 \right)}{h \left( \sqrt{9 + 2h} + 3 \right)}
   \]

   Simplifying the numerator using the difference of squares:

   \[
   f'(4) = \lim_{h \to 0} \frac{(9 + 2h) - 9}{h \left( \sqrt{9 + 2h} + 3 \right)} = \lim_{h \to 0} \frac{2h}{h \left( \sqrt{9 + 2h} + 3 \right)}
   \]

5. **Cancel out \( h \) from the numerator and denominator:**
   \[
   f'(4) = \lim_{h \to 0} \frac{2}{\sqrt{9 + 2h} + 3}
   \]

6. **Evaluate the limit as \( h \to 0 \):**
   As \( h \to 0 \), \( \sqrt{9 + 2h} \to \sqrt{9} = 3 \), so:

   \[
   f'(4) = \frac{2}{3 + 3} = \frac{2}{6} = \frac{1}{3}
   \]

Thus, the derivative of \( f(x) = \sqrt{2x + 1} \) at \( x = 4 \) is:

\[
f'(4) = \frac{1}{3}
\]

---

Would you like a more detailed explanation of any step or have any other questions?

### Related Questions:
1. How can you derive the formula for the derivative of \( f(x) = \sqrt{2x+1} \) at any general point \( x \)?
2. What is the significance of using the limit definition for finding derivatives?
3. Can you apply the same limit definition to find the derivative of \( f(x) = \sqrt{x^2 + 1} \)?
4. What are some alternative methods to find the derivative of a square root function?
5. How does multiplying by the conjugate help in simplifying the limit expression?

### Tip:
When applying the limit definition of the derivative, rationalizing the expression by using conjugates is often a key step for functions involving square roots.

Using the definition of the derivative (and not derivative rules), find the derivative of the function \( \sqrt{2x + 1} \) at \( x = 4 \). You will not receive credit if you do not use the limit definition of the derivative. Be sure to use proper limit notation as you are showing your work.

This problem uses the limit definition of the derivative to find the derivative of the square root function \( f(x) = \sqrt{2x + 1} \) at \( x = 4 \). Step-by-step solution included, suitable for high school calculus students.

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