Let's go through the problem step by step to compute $h'(2)$ for each function $h(x)$, given the information:

• $f'(2) = 12$
• $g'(2) = -1$

(A) $h(x) = 4f(x)$

Using the chain rule: $h'(x) = 4f'(x)$ Thus, $h'(2) = 4 \cdot f'(2) = 4 \cdot 12 = 48$.

(B) $h(x) = -13g(x)$

$h'(x) = -13g'(x)$ Thus, $h'(2) = -13 \cdot g'(2) = -13 \cdot (-1) = 13$.

(C) $h(x) = 10f(x) + 13g(x)$

$h'(x) = 10f'(x) + 13g'(x)$ Thus, $h'(2) = 10 \cdot f'(2) + 13 \cdot g'(2) = 10 \cdot 12 + 13 \cdot (-1) = 120 - 13 = 107$.

(D) $h(x) = 12g(x) - 7f(x)$

$h'(x) = 12g'(x) - 7f'(x)$ Thus, $h'(2) = 12 \cdot g'(2) - 7 \cdot f'(2) = 12 \cdot (-1) - 7 \cdot 12 = -12 - 84 = -96$.

(E) $h(x) = 8f(x) + 3g(x) - 2$

The derivative of the constant $-2$ is zero: $h'(x) = 8f'(x) + 3g'(x)$ Thus, $h'(2) = 8 \cdot f'(2) + 3 \cdot g'(2) = 8 \cdot 12 + 3 \cdot (-1) = 96 - 3 = 93$.

(F) $h(x) = -8g(x) - 7f(x) - 10x$

The derivative of $-10x$ is $-10$: $h'(x) = -8g'(x) - 7f'(x) - 10$ Thus, $h'(2) = -8 \cdot g'(2) - 7 \cdot f'(2) - 10 = -8 \cdot (-1) - 7 \cdot 12 - 10 = 8 - 84 - 10 = -86$.

If you have any questions or need further details, feel free to ask!

Follow-up questions:

1. What is the significance of using the chain rule in these derivative calculations?
2. How would the answers change if the derivatives $f'(2)$ and $g'(2)$ had different values?
3. Why does the derivative of a constant term, like in (E), not affect the result?
4. Can you explain why $f'(2)$ and $g'(2)$ are essential for these calculations?
5. How do different coefficients (like the 4 in $4f(x)$) affect the derivative of $h(x)$?

Tip:

When dealing with combinations of functions in derivative problems, always apply the chain rule and product rule carefully to avoid mistakes.