1. **Problem:** Find the derivative of the function $f(x) = (2x^2 - 3x)(-4x^{-2} + 5)$. 2. **Formula and rules:** Use the product rule for derivatives: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ where $u(x) = 2x^2 - 3x$ and $v(x) = -4x^{-2} + 5$. 3. **Find derivatives of $u(x)$ and $v(x)$:** $$u'(x) = \frac{d}{dx}(2x^2 - 3x) = 4x - 3$$ $$v'(x) = \frac{d}{dx}(-4x^{-2} + 5) = 8x^{-3}$$ 4. **Apply product rule:** $$f'(x) = (4x - 3)(-4x^{-2} + 5) + (2x^2 - 3x)(8x^{-3})$$ 5. **Expand terms:** $$= (4x)(-4x^{-2}) + (4x)(5) - 3(-4x^{-2}) - 3(5) + (2x^2)(8x^{-3}) - (3x)(8x^{-3})$$ $$= -16x^{-1} + 20x + 12x^{-2} - 15 + 16x^{-1} - 24x^{-2}$$ 6. **Combine like terms:** $$(-16x^{-1} + 16x^{-1}) + (12x^{-2} - 24x^{-2}) + 20x - 15$$ $$= 0 - 12x^{-2} + 20x - 15$$ 7. **Final answer:** $$\boxed{f'(x) = 20x - 12x^{-2} - 15}$$