Derivative of (x^3-4x+1)^3

1. Let $u = \left(x^{3} - 4 x\right) + 1$.

2. Apply the power rule: $u^{3}$ goes to $3 u^{2}$

3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(\left(x^{3} - 4 x\right) + 1\right)$:

   1. Differentiate $\left(x^{3} - 4 x\right) + 1$ term by term:

      1. Differentiate $x^{3} - 4 x$ term by term:

         1. Apply the power rule: $x^{3}$ goes to $3 x^{2}$

         2. The derivative of a constant times a function is the constant times the derivative of the function.

            1. Apply the power rule: $x$ goes to $1$

            So, the result is: $-4$

         The result is: $3 x^{2} - 4$

      2. The derivative of the constant $1$ is zero.

      The result is: $3 x^{2} - 4$

   The result of the chain rule is:

   $$3 \left(3 x^{2} - 4\right) \left(\left(x^{3} - 4 x\right) + 1\right)^{2}$$

4. Now simplify:

   $$\left(9 x^{2} - 12\right) \left(x^{3} - 4 x + 1\right)^{2}$$

The answer is:

$$\left(9 x^{2} - 12\right) \left(x^{3} - 4 x + 1\right)^{2}$$