**Problem:** Differentiate the function $$f(x) = (x^2 - x)(2x^3 - 1)^3$$. 1. **State the problem:** We need to find the derivative $$f'(x)$$ of the product of two functions: $$u(x) = x^2 - x$$ and $$v(x) = (2x^3 - 1)^3$$. 2. **Recall the product rule:** $$ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) $$ 3. **Find derivatives of each part:** - Derivative of $$u(x) = x^2 - x$$ is $$ u'(x) = 2x - 1 $$ - To find $$v'(x)$$ where $$v(x) = (2x^3 - 1)^3$$, use the chain rule: - Let $$g(x) = 2x^3 - 1$$, then $$v(x) = [g(x)]^3$$. - Derivative of $$v(x)$$ is $$ v'(x) = 3[g(x)]^2 \cdot g'(x) $$ - Derivative of $$g(x) = 2x^3 - 1$$ is $$ g'(x) = 6x^2 $$ - So, $$ v'(x) = 3(2x^3 - 1)^2 \cdot 6x^2 = 18x^2 (2x^3 - 1)^2 $$ 4. **Apply the product rule:** $$ f'(x) = u'(x)v(x) + u(x)v'(x) = (2x - 1)(2x^3 - 1)^3 + (x^2 - x) \cdot 18x^2 (2x^3 - 1)^2 $$ 5. **Final answer:** $$ f'(x) = (2x - 1)(2x^3 - 1)^3 + 18x^2 (x^2 - x)(2x^3 - 1)^2 $$ This is the derivative of the given function using the product and chain rules.