1. **Problem Statement:** Given values $f(-3) = -8$, $g(-3) = 6$, $f'(-3) = 2$, and $g'(-3) = 8$, find $h'(-3)$ for each function $h(x)$. 2. **Recall the derivative rules:** - Constant multiple rule: $\frac{d}{dx}[cf(x)] = c f'(x)$ - Sum/difference rule: $\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$ - Product rule: $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$ - Quotient rule: $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$ --- **(A) For** $h(x) = 5f(x) - 4g(x)$: 3. Differentiate using constant multiple and difference rules: $$h'(x) = 5f'(x) - 4g'(x)$$ 4. Substitute $x = -3$: $$h'(-3) = 5 \times 2 - 4 \times 8 = 10 - 32 = -22$$ --- **(B) For** $h(x) = f(x)g(x)$: 5. Use product rule: $$h'(x) = f'(x)g(x) + f(x)g'(x)$$ 6. Substitute $x = -3$: $$h'(-3) = 2 \times 6 + (-8) \times 8 = 12 - 64 = -52$$ --- **(C) For** $h(x) = \frac{f(x)}{g(x)}$: 7. Use quotient rule: $$h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$$ 8. Substitute $x = -3$: $$h'(-3) = \frac{2 \times 6 - (-8) \times 8}{6^2} = \frac{12 + 64}{36} = \frac{76}{36}$$ 9. Simplify fraction: $$\frac{76}{36} = \frac{\cancel{4} \times 19}{\cancel{4} \times 9} = \frac{19}{9}$$ --- **(D) For** $h(x) = \frac{g(x)}{1 + f(x)}$: 10. Use quotient rule with numerator $g(x)$ and denominator $1 + f(x)$: $$h'(x) = \frac{g'(x)(1 + f(x)) - g(x)f'(x)}{(1 + f(x))^2}$$ 11. Substitute $x = -3$: $$h'(-3) = \frac{8(1 - 8) - 6 \times 2}{(1 - 8)^2} = \frac{8 \times (-7) - 12}{(-7)^2} = \frac{-56 - 12}{49} = \frac{-68}{49}$$ --- **Final answers:** - (A) $h'(-3) = -22$ - (B) $h'(-3) = -52$ - (C) $h'(-3) = \frac{19}{9}$ - (D) $h'(-3) = \frac{-68}{49}$