https://file.aiquickdraw.com/mathbot/file/17274521630796ook6tji.jpg

Let's solve the provided problems using the table of values.

### 1. \((f \cdot g)(0)\)
This is the product of \(f(0)\) and \(g(0)\):
\[
(f \cdot g)(0) = f(0) \cdot g(0) = 2 \cdot 1 = 2
\]

### 2. \(h'(0)\) where \(h(x) = f(g(x))\)
To find \(h'(x)\), use the chain rule:
\[
h'(x) = f'(g(x)) \cdot g'(x)
\]
At \(x = 0\):
\[
h'(0) = f'(g(0)) \cdot g'(0) = f'(1) \cdot g'(0)
\]
From the table:
- \(f'(1) = 4\)
- \(g'(0) = 2\)
Thus:
\[
h'(0) = 4 \cdot 2 = 8
\]

### 3. \(f(f(0))\)
We first find \(f(0)\) and then apply \(f\) again:
\[
f(0) = 2 \quad \text{and} \quad f(2) = 10
\]
So:
\[
f(f(0)) = f(2) = 10
\]

### 4. \((\frac{f}{g})'(-2)\)
For the derivative of \(\frac{f}{g}\), use the quotient rule:
\[
\left( \frac{f}{g} \right)'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}
\]
At \(x = -2\):
\[
\left( \frac{f}{g} \right)'(-2) = \frac{f'(-2)g(-2) - f(-2)g'(-2)}{(g(-2))^2}
\]
From the table:
- \(f'(-2) = -2\)
- \(g(-2) = 1\)
- \(f(-2) = 6\)
- \(g'(-2) = -2\)
Thus:
\[
\left( \frac{f}{g} \right)'(-2) = \frac{(-2)(1) - (6)(-2)}{(1)^2} = \frac{-2 + 12}{1} = 10
\]

### 5. \((f \cdot g)'(-2)\)
For the derivative of \(f \cdot g\), use the product rule:
\[
(f \cdot g)'(x) = f'(x)g(x) + f(x)g'(x)
\]
At \(x = -2\):
\[
(f \cdot g)'(-2) = f'(-2)g(-2) + f(-2)g'(-2)
\]
Substitute the values from the table:
- \(f'(-2) = -2\)
- \(g(-2) = 1\)
- \(f(-2) = 6\)
- \(g'(-2) = -2\)
Thus:
\[
(f \cdot g)'(-2) = (-2)(1) + (6)(-2) = -2 - 12 = -14
\]

### 6. \((f - g)'(-2)\)
For the derivative of \(f - g\):
\[
(f - g)'(x) = f'(x) - g'(x)
\]
At \(x = -2\):
\[
(f - g)'(-2) = f'(-2) - g'(-2)
\]
From the table:
- \(f'(-2) = -2\)
- \(g'(-2) = -2\)
Thus:
\[
(f - g)'(-2) = -2 - (-2) = 0
\]

### Final answers:
1. \((f \cdot g)(0) = 2\)
2. \(h'(0) = 8\)
3. \(f(f(0)) = 10\)
4. \((\frac{f}{g})'(-2) = 10\)
5. \((f \cdot g)'(-2) = -14\)
6. \((f - g)'(-2) = 0\)

Let me know if you want further details!

### Related Questions:
1. How is the product rule applied to functions like \(f \cdot g\)?
2. What is the chain rule, and how does it apply to composite functions?
3. How is the quotient rule different from the product rule?
4. Why does \((f - g)'(-2)\) result in zero?
5. What happens if either \(f(x)\) or \(g(x)\) is zero at a point?

### Tip:
When dealing with multiple derivatives, always check which rule (chain, product, or quotient) applies to avoid errors!

Use the table to calculate values including (f · g)(0), h'(0) where h(x) = f(g(x)), f(f(0)), (f/g)'(-2), (f · g)'(-2), and (f - g)'(-2).

Learn how to calculate derivatives using product, chain, and quotient rules with a step-by-step solution. This problem covers essential calculus concepts, including function composition and differentiation. Suitable for high school calculus students.

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