1. **State the problem:** Find the second derivative of the function $$y = x^2 e^{2x}$$. 2. **Recall the product rule:** For two functions $u(x)$ and $v(x)$, the derivative is $$\frac{d}{dx}[uv] = u'v + uv'$$. 3. **Identify parts:** Let $$u = x^2$$ and $$v = e^{2x}$$. 4. **Find first derivatives:** - $$u' = \frac{d}{dx} x^2 = 2x$$ - $$v' = \frac{d}{dx} e^{2x} = 2e^{2x}$$ (using chain rule). 5. **Apply product rule for first derivative:** $$y' = u'v + uv' = 2x e^{2x} + x^2 (2 e^{2x}) = 2x e^{2x} + 2x^2 e^{2x}$$ 6. **Simplify first derivative:** $$y' = 2x e^{2x} + 2x^2 e^{2x} = 2 e^{2x} (x + x^2) = 2 e^{2x} x (1 + x)$$ 7. **Find second derivative:** Differentiate $$y' = 2 e^{2x} x (1 + x)$$. Let $$w = 2 e^{2x}$$ and $$z = x (1 + x) = x + x^2$$. 8. **Find derivatives:** - $$w' = 2 \cdot 2 e^{2x} = 4 e^{2x}$$ - $$z' = \frac{d}{dx} (x + x^2) = 1 + 2x$$ 9. **Apply product rule again:** $$y'' = w' z + w z' = 4 e^{2x} (x + x^2) + 2 e^{2x} (1 + 2x)$$ 10. **Factor out $$2 e^{2x}$$:** $$y'' = 2 e^{2x} [2 (x + x^2) + (1 + 2x)] = 2 e^{2x} (2x + 2x^2 + 1 + 2x)$$ 11. **Combine like terms:** $$2x + 2x = 4x$$ So, $$y'' = 2 e^{2x} (1 + 4x + 2x^2)$$ **Final answer:** $$\boxed{y'' = 2 e^{2x} (1 + 4x + 2x^2)}$$