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