To find $f'(a)$ using the definition of a derivative, we apply the limit definition of the derivative:

$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

In this case, the function is $f(x) = 3x + 1$ and $a = 5$. Let's calculate step-by-step:

Step 1: Calculate $f(a)$

$f(a) = f(5) = 3(5) + 1 = 15 + 1 = 16$

Step 2: Calculate $f(a+h)$

$f(a+h) = f(5+h) = 3(5+h) + 1 = 15 + 3h + 1 = 16 + 3h$

Step 3: Apply the limit definition of the derivative

Now substitute these values into the derivative formula:

$f'(5) = \lim_{h \to 0} \frac{(16 + 3h) - 16}{h} = \lim_{h \to 0} \frac{3h}{h}$

Step 4: Simplify the expression

$f'(5) = \lim_{h \to 0} 3 = 3$

Thus, the derivative $f'(5) = 3$.

Would you like more details or have any questions about this? Here are some related questions for further understanding:

1. What is the general derivative of the function $f(x) = 3x + 1$?
2. How is the derivative related to the slope of the function?
3. How can the limit definition of a derivative be applied to more complex functions?
4. What is the geometric interpretation of $f'(a)$ for linear functions?
5. What happens to the derivative if we change $a$ to a different value?

Tip: The derivative of a linear function $f(x) = mx + b$ is always the slope $m$.