1. **Problem:** Differentiate the function $$y = (2x^2 + 3)^4 (x^4 - 3)^3$$. 2. **Formula and rules:** Use the product rule: $$\frac{d}{dx}[u v] = u' v + u v'$$ where $$u = (2x^2 + 3)^4$$ and $$v = (x^4 - 3)^3$$. Also use the chain rule for each term: $$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$. 3. **Differentiate each part:** - For $$u = (2x^2 + 3)^4$$, let $$f(t) = t^4$$ and $$g(x) = 2x^2 + 3$$. Then $$u' = 4(2x^2 + 3)^3 \cdot (4x) = 16x (2x^2 + 3)^3$$. - For $$v = (x^4 - 3)^3$$, let $$h(t) = t^3$$ and $$k(x) = x^4 - 3$$. Then $$v' = 3(x^4 - 3)^2 \cdot 4x^3 = 12x^3 (x^4 - 3)^2$$. 4. **Apply product rule:** $$\frac{dy}{dx} = u' v + u v' = 16x (2x^2 + 3)^3 (x^4 - 3)^3 + (2x^2 + 3)^4 12x^3 (x^4 - 3)^2$$. 5. **Final answer:** $$\boxed{\frac{dy}{dx} = 16x (2x^2 + 3)^3 (x^4 - 3)^3 + 12x^3 (2x^2 + 3)^4 (x^4 - 3)^2}$$