1. Problem: Find the derivative of the function $ f(x) = \left( 2x^2 - 3x \right) \left( -4x^{-2} + 5 \right) $. 2. Formula: Use the product rule for derivatives: $ (uv)' = u'v + uv' $. 3. Let $ u = 2x^2 - 3x $ and $ v = -4x^{-2} + 5 $. 4. Compute derivatives: $ u' = \frac{d}{dx}(2x^2 - 3x) = 4x - 3 $ $ v' = \frac{d}{dx}(-4x^{-2} + 5) = 8x^{-3} $ 5. Apply product rule: $$ f'(x) = u'v + uv' = (4x - 3)(-4x^{-2} + 5) + (2x^2 - 3x)(8x^{-3}) $$ 6. Expand terms: $$ (4x - 3)(-4x^{-2} + 5) = 4x \cdot (-4x^{-2}) + 4x \cdot 5 - 3 \cdot (-4x^{-2}) - 3 \cdot 5 $$ $$ = -16x^{-1} + 20x + 12x^{-2} - 15 $$ $$ (2x^2 - 3x)(8x^{-3}) = 2x^2 \cdot 8x^{-3} - 3x \cdot 8x^{-3} = 16x^{-1} - 24x^{-2} $$ 7. Combine all terms: $$ f'(x) = (-16x^{-1} + 20x + 12x^{-2} - 15) + (16x^{-1} - 24x^{-2}) $$ 8. Simplify by canceling terms: $$ -16x^{-1} + 16x^{-1} = \cancel{-16x^{-1} + 16x^{-1}} = 0 $$ $$ 12x^{-2} - 24x^{-2} = \cancel{12x^{-2} - 24x^{-2}} = -12x^{-2} $$ 9. Final simplified derivative: $$ f'(x) = 20x - 15 - 12x^{-2} $$ Answer: $ \boxed{f'(x) = 20x - 15 - 12x^{-2}} $