1. **State the problem:** Find the derivative of the function $$y = x(2 - e^x)^3$$. 2. **Formula used:** We will use the product rule and the chain rule. - Product rule: $$\frac{d}{dx}[u \cdot v] = u'v + uv'$$ - Chain rule: $$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$ 3. **Identify parts:** Let $$u = x$$ and $$v = (2 - e^x)^3$$. 4. **Differentiate each part:** - $$u' = \frac{d}{dx}[x] = 1$$ - For $$v'$$, use chain rule: - Outer function: $$f(t) = t^3$$, so $$f'(t) = 3t^2$$ - Inner function: $$g(x) = 2 - e^x$$, so $$g'(x) = -e^x$$ - Therefore, $$v' = 3(2 - e^x)^2 \cdot (-e^x) = -3e^x(2 - e^x)^2$$ 5. **Apply product rule:** $$\frac{dy}{dx} = u'v + uv' = 1 \cdot (2 - e^x)^3 + x \cdot \left(-3e^x(2 - e^x)^2\right)$$ 6. **Simplify:** $$\frac{dy}{dx} = (2 - e^x)^3 - 3xe^x(2 - e^x)^2$$ 7. **Factor common terms:** $$\frac{dy}{dx} = (2 - e^x)^2 \left( (2 - e^x) - 3xe^x \right)$$ **Final answer:** $$\boxed{\frac{dy}{dx} = (2 - e^x)^2 \left( 2 - e^x - 3xe^x \right)}$$