(1 point)
Find the derivative of the vector function
$\mathbf{r}(t)=t \mathbf{a} \times(\mathbf{b}+t \mathbf{c})$, where
$\mathbf{a}=\langle 1,-1,1\rangle, \mathbf{b}=\langle 4,-4,-4\rangle$, and $\mathbf{c}=\langle-2,1,-4\rangle$
$\mathbf{r}^{\prime}(t)=\langle$

Step 1 ：We are given the vector function \(\mathbf{r}(t)=t \mathbf{a} \times(\mathbf{b}+t \mathbf{c})\), where \(\mathbf{a}=\langle 1,-1,1\rangle\), \(\mathbf{b}=\langle 4,-4,-4\rangle\), and \(\mathbf{c}=\langle-2,1,-4\rangle\).

Step 2 ：The derivative of a vector function is found by taking the derivative of each of its components separately. In this case, we have a cross product involved, which complicates things a bit.

Step 3 ：The cross product of two vectors a and b is given by: \(\mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2)i - (a_1b_3 - a_3b_1)j + (a_1b_2 - a_2b_1)k\), where \(a_1, a_2, a_3\) and \(b_1, b_2, b_3\) are the components of vectors a and b respectively, and i, j, k are the unit vectors in the x, y, z directions.

Step 4 ：The derivative of a cross product follows the product rule, which states that the derivative of a product of two functions is the derivative of the first times the second plus the first times the derivative of the second.

Step 5 ：So, we need to find the derivative of \(t\mathbf{a}\) and \(\mathbf{b} + t\mathbf{c}\) separately, and then apply the product rule.

Step 6 ：By applying the product rule and simplifying, we find that the derivative of the vector function \(\mathbf{r}(t)=t \mathbf{a} \times(\mathbf{b}+t \mathbf{c})\) is \(\mathbf{r}^\prime(t)=\langle 6t + 8, 4t + 8, -2t \rangle\).

Step 7 ：Final Answer: \(\boxed{\mathbf{r}^\prime(t)=\langle 6t + 8, 4t + 8, -2t \rangle}\)