1. The problem is to find the function $H_4(x)$ defined by the formula $$H_4(x) = (-1)^4 e^{x^2} \frac{d^4}{dx^4} e^{-x^2}.$$ 2. This formula involves taking the fourth derivative of the function $e^{-x^2}$, then multiplying by $e^{x^2}$ and $(-1)^4$. 3. Note that $(-1)^4 = 1$, so the formula simplifies to $$H_4(x) = e^{x^2} \frac{d^4}{dx^4} e^{-x^2}.$$ 4. We will compute the derivatives step-by-step. First, let $$f(x) = e^{-x^2}.$$ 5. Compute the first derivative: $$f'(x) = \frac{d}{dx} e^{-x^2} = -2x e^{-x^2}.$$ 6. Compute the second derivative: $$f''(x) = \frac{d}{dx} (-2x e^{-x^2}) = -2 e^{-x^2} + 4x^2 e^{-x^2} = (4x^2 - 2) e^{-x^2}.$$ 7. Compute the third derivative: $$f'''(x) = \frac{d}{dx} ((4x^2 - 2) e^{-x^2}) = (8x) e^{-x^2} + (4x^2 - 2)(-2x) e^{-x^2} = (8x - 8x^3 + 4x) e^{-x^2} = (-8x^3 + 12x) e^{-x^2}.$$ 8. Compute the fourth derivative: $$f^{(4)}(x) = \frac{d}{dx}((-8x^3 + 12x) e^{-x^2}) = (-24x^2 + 12) e^{-x^2} + (-8x^3 + 12x)(-2x) e^{-x^2} = (-24x^2 + 12 + 16x^4 - 24x^2) e^{-x^2} = (16x^4 - 48x^2 + 12) e^{-x^2}.$$ 9. Substitute back into the formula for $H_4(x)$: $$H_4(x) = e^{x^2} \times (16x^4 - 48x^2 + 12) e^{-x^2} = 16x^4 - 48x^2 + 12.$$ 10. Therefore, the final answer is $$\boxed{H_4(x) = 16x^4 - 48x^2 + 12}.$$