1. **State the problem:** Differentiate the function $$f(x) = (2x^3 + 5)(3x^2 + 5x)$$ with respect to $$x$$. 2. **Formula used:** Use the product rule for differentiation, which states: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ where $$u(x) = 2x^3 + 5$$ and $$v(x) = 3x^2 + 5x$$. 3. **Find the derivatives of each function:** $$u'(x) = \frac{d}{dx}(2x^3 + 5) = 6x^2$$ $$v'(x) = \frac{d}{dx}(3x^2 + 5x) = 6x + 5$$ 4. **Apply the product rule:** $$f'(x) = u'(x)v(x) + u(x)v'(x) = 6x^2(3x^2 + 5x) + (2x^3 + 5)(6x + 5)$$ 5. **Expand each term:** $$6x^2(3x^2 + 5x) = 18x^4 + 30x^3$$ $$ (2x^3 + 5)(6x + 5) = 2x^3 \times 6x + 2x^3 \times 5 + 5 \times 6x + 5 \times 5 = 12x^4 + 10x^3 + 30x + 25$$ 6. **Combine all terms:** $$f'(x) = (18x^4 + 30x^3) + (12x^4 + 10x^3 + 30x + 25) = (18x^4 + 12x^4) + (30x^3 + 10x^3) + 30x + 25 = 30x^4 + 40x^3 + 30x + 25$$ **Final answer:** $$\boxed{f'(x) = 30x^4 + 40x^3 + 30x + 25}$$