To find $f'(a)$ for $f(x) = \sqrt{2x + 1}$ at $a = 2$, we need to differentiate the function and evaluate the derivative at $x = 2$.

Step-by-Step Solution

1. Rewrite the function in a form that makes differentiation easier: $f(x) = (2x + 1)^{1/2}$

2. Differentiate $f(x)$ with respect to $x$: Using the chain rule, where $f(x) = (u)^{1/2}$ and $u = 2x + 1$, $f'(x) = \frac{1}{2} (2x + 1)^{-1/2} \cdot (2)$

3. Simplify the derivative: $f'(x) = \frac{1}{2} \cdot \frac{2}{\sqrt{2x + 1}} = \frac{1}{\sqrt{2x + 1}}$

4. Evaluate $f'(x)$ at $x = 2$: Substitute $x = 2$ into the derivative: $f'(2) = \frac{1}{\sqrt{2(2) + 1}} = \frac{1}{\sqrt{4 + 1}} = \frac{1}{\sqrt{5}}$

Thus, the answer is: $f'(2) = \frac{1}{\sqrt{5}}$

Would you like further details on any step, or have any questions?

Here are some related questions for practice:

1. What is the derivative of $f(x) = \sqrt{3x + 2}$ at $x = 1$?
2. How would you find $f'(a)$ if $f(x) = \sqrt{5x + 4}$ at $a = 3$?
3. How do you apply the chain rule when differentiating $f(x) = \sqrt{7x + 1}$?
4. What is the derivative of $g(x) = x^{1/2}$ at $x = 9$?
5. How would you evaluate $f'(x)$ at $x = 0$ for $f(x) = \sqrt{2x + 3}$?

Tip: For functions of the form $f(x) = (ax + b)^{n}$, remember to apply the chain rule to differentiate effectively.