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!