1. **Problem statement:** Given the function $f(x) = x^3 - 4x^2 - 4x - 1$, find the first derivative $f'(x)$. 2. **Formula and rules:** The derivative of a function $f(x)$ with respect to $x$ is found by applying the power rule: $$\frac{d}{dx} x^n = n x^{n-1}$$ and the sum rule: $$\frac{d}{dx} (u + v) = \frac{du}{dx} + \frac{dv}{dx}$$. 3. **Step-by-step differentiation:** - Derivative of $x^3$ is $3x^2$. - Derivative of $-4x^2$ is $-4 \times 2x = -8x$. - Derivative of $-4x$ is $-4$. - Derivative of constant $-1$ is $0$. 4. **Combine all:** $$f'(x) = 3x^2 - 8x - 4$$ **Final answer:** $$\boxed{f'(x) = 3x^2 - 8x - 4}$$