1. **State the problem:** We need to find the derivative $\frac{dy}{dt}$ of the function $$y = (t^2 + 6t + 6)(4t^2 + 3).$$ 2. **Formula used:** To differentiate a product of two functions, use the product rule: $$\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t),$$ where $u(t) = t^2 + 6t + 6$ and $v(t) = 4t^2 + 3$. 3. **Find derivatives of each function:** $$u'(t) = \frac{d}{dt}(t^2 + 6t + 6) = 2t + 6,$$ $$v'(t) = \frac{d}{dt}(4t^2 + 3) = 8t.$$ 4. **Apply the product rule:** $$\frac{dy}{dt} = u'(t)v(t) + u(t)v'(t) = (2t + 6)(4t^2 + 3) + (t^2 + 6t + 6)(8t).$$ 5. **Expand each term:** $$(2t + 6)(4t^2 + 3) = 2t \cdot 4t^2 + 2t \cdot 3 + 6 \cdot 4t^2 + 6 \cdot 3 = 8t^3 + 6t + 24t^2 + 18,$$ $$(t^2 + 6t + 6)(8t) = 8t \cdot t^2 + 8t \cdot 6t + 8t \cdot 6 = 8t^3 + 48t^2 + 48t.$$ 6. **Combine like terms:** $$\frac{dy}{dt} = (8t^3 + 24t^2 + 6t + 18) + (8t^3 + 48t^2 + 48t) = (8t^3 + 8t^3) + (24t^2 + 48t^2) + (6t + 48t) + 18 = 16t^3 + 72t^2 + 54t + 18.$$ **Final answer:** $$\frac{dy}{dt} = 16t^3 + 72t^2 + 54t + 18.$$