1. **State the problem:** Differentiate the function $$f(x) = \frac{4}{(5-x)(x-1)}$$ with respect to $$x$$. 2. **Rewrite the function:** To differentiate, it's easier to write $$f(x)$$ as $$4 \cdot \frac{1}{(5-x)(x-1)}$$. 3. **Use the quotient/product rule:** Since the denominator is a product, let $$g(x) = (5-x)(x-1)$$. Then $$f(x) = \frac{4}{g(x)} = 4g(x)^{-1}$$. 4. **Differentiate using the chain rule:** $$f'(x) = 4 \cdot (-1) g(x)^{-2} \cdot g'(x) = -4 \frac{g'(x)}{(g(x))^2}$$ 5. **Find $$g'(x)$$:** $$g(x) = (5-x)(x-1)$$ Use the product rule: $$g'(x) = (5-x)'(x-1) + (5-x)(x-1)' = (-1)(x-1) + (5-x)(1) = - (x-1) + (5-x)$$ Simplify: $$-x + 1 + 5 - x = 6 - 2x$$ 6. **Substitute back:** $$f'(x) = -4 \frac{6 - 2x}{[(5-x)(x-1)]^2}$$ 7. **Final answer:** $$\boxed{f'(x) = -4 \frac{6 - 2x}{[(5-x)(x-1)]^2}}$$ This derivative tells us the rate of change of the function $$f(x)$$ at any point $$x$$, considering the product in the denominator and applying the chain and product rules carefully.