Let's solve the problem step by step.

Given Function:

$f(x) = \sqrt{7x + 1}$ You are asked to:

1. Find $f'(a)$ when $a = 5$.

Step 1: Find the derivative of $f(x)$.

We will use the chain rule to differentiate $f(x) = \sqrt{7x + 1} = (7x + 1)^{1/2}$.

Using the chain rule: $f'(x) = \frac{1}{2} (7x + 1)^{-1/2} \cdot 7$ Simplifying: $f'(x) = \frac{7}{2\sqrt{7x + 1}}$

Step 2: Evaluate $f'(a)$ when $a = 5$.

Substitute $x = 5$ into the derivative: $f'(5) = \frac{7}{2\sqrt{7(5) + 1}} = \frac{7}{2\sqrt{36}} = \frac{7}{2 \times 6} = \frac{7}{12}$

So, $f'(5) = \frac{7}{12}$.

Would you like more details or help with part b of the problem? Here are some follow-up questions for deeper understanding:

1. Can you describe how the chain rule was applied in this problem?
2. Why is it important to simplify the expression for $f'(x)$ before plugging in values?
3. What is the geometrical interpretation of $f'(a)$?
4. How does the derivative relate to the slope of the tangent line at a point on a graph?
5. Could you explain how $f'(a)$ helps find the equation of the tangent line?

Tip: Always check if functions like square roots can be simplified before applying differentiation rules to reduce complexity.