1. Stating the problem: Find the derivative of the function $$y = (5 - 2x)^{-3} + \frac{8}{1} (x^2 + 1)^4 = (5 - 2x)^{-3} + 8(x^2 + 1)^4$$. 2. Rewrite the function for clarity: $$y = (5 - 2x)^{-3} + 8(x^2 + 1)^4$$. 3. Differentiate the first term using the chain rule: The outer function is $$u^{-3}$$ where $$u = 5 - 2x$$. Derivative of $$u^{-3}$$ with respect to $$u$$ is $$-3u^{-4}$$. Derivative of $$u$$ with respect to $$x$$ is $$-2$$. So, $$\frac{d}{dx} (5-2x)^{-3} = -3(5 - 2x)^{-4} \cdot (-2) = 6(5 - 2x)^{-4}$$. 4. Differentiate the second term using the chain rule: The function is $$8 (x^2 + 1)^4$$. The outer function is $$v^4$$ where $$v = x^2 + 1$$. Derivative of $$v^4$$ with respect to $$v$$ is $$4v^3$$. Derivative of $$v$$ with respect to $$x$$ is $$2x$$. So, $$\frac{d}{dx} 8(x^2+1)^4 = 8 \cdot 4 (x^2 + 1)^3 \cdot 2x = 64x (x^2 + 1)^3$$. 5. Combine the derivatives: $$\frac{dy}{dx} = 6(5 - 2x)^{-4} + 64x (x^2 + 1)^3$$. Final answer: $$\boxed{\frac{dy}{dx} = 6(5 - 2x)^{-4} + 64x (x^2 + 1)^3}$$