1. Problem: Find the derivative of the function $$f(x) = 3x^2 + 3\sqrt[3]{x} - \frac{2}{x^3}$$ 2. Formula and rules: - Power rule: $$\frac{d}{dx} x^n = nx^{n-1}$$ - Derivative of a constant times a function: $$\frac{d}{dx} [cf(x)] = c f'(x)$$ - Derivative of a sum: $$\frac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x)$$ - Rewrite roots and fractions as powers: $$\sqrt[3]{x} = x^{1/3}$$ and $$\frac{1}{x^3} = x^{-3}$$ 3. Apply the power rule to each term: - $$\frac{d}{dx} 3x^2 = 3 \cdot 2x^{2-1} = 6x$$ - $$\frac{d}{dx} 3x^{1/3} = 3 \cdot \frac{1}{3} x^{1/3 - 1} = x^{-2/3}$$ - $$\frac{d}{dx} \left(-2x^{-3}\right) = -2 \cdot (-3) x^{-3 - 1} = 6x^{-4}$$ 4. Combine the derivatives: $$f'(x) = 6x + x^{-2/3} + 6x^{-4}$$ Final answer: $$\boxed{f'(x) = 6x + x^{-2/3} + 6x^{-4}}$$