1. **State the problem:** We need to find the third derivative $f'''(x)$ of the function $$f(x) = -x^2 - 2x^3 + 5x^4 + e^{-4x}$$ and then evaluate it at $x=0$. 2. **Recall the rules for derivatives:** - The derivative of $x^n$ is $nx^{n-1}$. - The derivative of $e^{g(x)}$ is $g'(x)e^{g(x)}$. 3. **Find the first derivative $f'(x)$:** $$f'(x) = \frac{d}{dx}(-x^2) + \frac{d}{dx}(-2x^3) + \frac{d}{dx}(5x^4) + \frac{d}{dx}(e^{-4x})$$ Calculate each term: - $\frac{d}{dx}(-x^2) = -2x$ - $\frac{d}{dx}(-2x^3) = -6x^2$ - $\frac{d}{dx}(5x^4) = 20x^3$ - $\frac{d}{dx}(e^{-4x}) = -4e^{-4x}$ (using chain rule) So, $$f'(x) = -2x - 6x^2 + 20x^3 - 4e^{-4x}$$ 4. **Find the second derivative $f''(x)$:** $$f''(x) = \frac{d}{dx}(-2x) + \frac{d}{dx}(-6x^2) + \frac{d}{dx}(20x^3) + \frac{d}{dx}(-4e^{-4x})$$ Calculate each term: - $\frac{d}{dx}(-2x) = -2$ - $\frac{d}{dx}(-6x^2) = -12x$ - $\frac{d}{dx}(20x^3) = 60x^2$ - $\frac{d}{dx}(-4e^{-4x}) = 16e^{-4x}$ (chain rule again) So, $$f''(x) = -2 - 12x + 60x^2 + 16e^{-4x}$$ 5. **Find the third derivative $f'''(x)$:** $$f'''(x) = \frac{d}{dx}(-2) + \frac{d}{dx}(-12x) + \frac{d}{dx}(60x^2) + \frac{d}{dx}(16e^{-4x})$$ Calculate each term: - $\frac{d}{dx}(-2) = 0$ - $\frac{d}{dx}(-12x) = -12$ - $\frac{d}{dx}(60x^2) = 120x$ - $\frac{d}{dx}(16e^{-4x}) = -64e^{-4x}$ So, $$f'''(x) = -12 + 120x - 64e^{-4x}$$ 6. **Evaluate $f'''(0)$:** Substitute $x=0$: $$f'''(0) = -12 + 120 \times 0 - 64e^{0} = -12 - 64 = -76$$ **Final answer:** $$\boxed{-76}$$