1. **State the problem:** Find the derivative of the function $$f(x) = \frac{4 - 5x}{2x^2 - 4x} \cdot (-x^4 + 3x - 2)(x^5 - 2x^4)$$ 2. **Rewrite the function:** Let $$g(x) = \frac{4 - 5x}{2x^2 - 4x}$$ and $$h(x) = (-x^4 + 3x - 2)(x^5 - 2x^4)$$ So, $$f(x) = g(x) \cdot h(x)$$ 3. **Use the product rule:** The derivative of a product is $$f'(x) = g'(x)h(x) + g(x)h'(x)$$ 4. **Find $g'(x)$:** Use the quotient rule for $$g(x) = \frac{u}{v} = \frac{4 - 5x}{2x^2 - 4x}$$ where $$u = 4 - 5x, \quad v = 2x^2 - 4x$$ The quotient rule is $$g'(x) = \frac{u'v - uv'}{v^2}$$ Calculate derivatives: $$u' = -5$$ $$v' = 4x - 4$$ So, $$g'(x) = \frac{(-5)(2x^2 - 4x) - (4 - 5x)(4x - 4)}{(2x^2 - 4x)^2}$$ Simplify numerator: $$-5(2x^2 - 4x) = -10x^2 + 20x$$ $$(4 - 5x)(4x - 4) = 16x - 16 - 20x^2 + 20x = -20x^2 + 36x - 16$$ So numerator: $$-10x^2 + 20x - (-20x^2 + 36x - 16) = -10x^2 + 20x + 20x^2 - 36x + 16 = 10x^2 - 16x + 16$$ Therefore, $$g'(x) = \frac{10x^2 - 16x + 16}{(2x^2 - 4x)^2}$$ 5. **Find $h'(x)$:** First expand $h(x)$: $$h(x) = (-x^4 + 3x - 2)(x^5 - 2x^4)$$ Multiply: $$-x^4 \cdot x^5 = -x^9$$ $$-x^4 \cdot (-2x^4) = +2x^8$$ $$3x \cdot x^5 = 3x^6$$ $$3x \cdot (-2x^4) = -6x^5$$ $$-2 \cdot x^5 = -2x^5$$ $$-2 \cdot (-2x^4) = +4x^4$$ So, $$h(x) = -x^9 + 2x^8 + 3x^6 - 6x^5 - 2x^5 + 4x^4 = -x^9 + 2x^8 + 3x^6 - 8x^5 + 4x^4$$ Now differentiate term by term: $$h'(x) = -9x^8 + 16x^7 + 18x^5 - 40x^4 + 16x^3$$ 6. **Write the final derivative:** $$f'(x) = g'(x)h(x) + g(x)h'(x) = \frac{10x^2 - 16x + 16}{(2x^2 - 4x)^2} \cdot (-x^9 + 2x^8 + 3x^6 - 8x^5 + 4x^4) + \frac{4 - 5x}{2x^2 - 4x} \cdot (-9x^8 + 16x^7 + 18x^5 - 40x^4 + 16x^3)$$ This is the derivative of the given function.