The problem asks to find the second derivative of the given functions. Let's go through the derivatives step by step.

1. (f) $f''(x)$ for $f(x) = \sqrt{9 - x}$

   The first derivative of $f(x) = \sqrt{9 - x}$ requires using the chain rule. First, rewrite the function as: $f(x) = (9 - x)^{1/2}$ Now, differentiate: $f'(x) = \frac{1}{2}(9 - x)^{-1/2} \cdot (-1)$ Simplifying: $f'(x) = -\frac{1}{2}(9 - x)^{-1/2}$ Now, to find the second derivative, we differentiate $f'(x)$ again: $f''(x) = -\frac{1}{2} \cdot (-1) \cdot (9 - x)^{-3/2} \cdot (-1)$ Simplifying: $f''(x) = \frac{1}{4}(9 - x)^{-3/2}$

2. (g) $y''$ for $y = (3x + 7)^5$

   First, apply the chain rule for the first derivative: $y' = 5(3x + 7)^4 \cdot 3 = 15(3x + 7)^4$ Now, differentiate again to find $y''$: $y'' = 15 \cdot 4(3x + 7)^3 \cdot 3 = 180(3x + 7)^3$

Let me know if you need further clarification or more steps! Would you like to continue with the other parts or need further explanations?

Here are 5 follow-up questions that may interest you:

1. How do we apply the chain rule in more complex functions?
2. How do you differentiate higher powers like $(3x + 7)^5$?
3. What are common mistakes when calculating higher-order derivatives?
4. How do we approach finding derivatives for trigonometric functions?
5. Can you find the third derivative of the same functions as a challenge?

Tip: Always double-check your chain rule applications, especially when differentiating composite functions.