14 Given $f(x)=x\left(1-3 x^{2}\right), f^{\prime}(-1)$ equals: \[ \begin{array}{ll} \text { A } & -8 \\ \text { B } & -5 \\ \text { C } & 2 \\ \text { D } & 6 \\ \text { E } & 10 \end{array} \]

Step 1 ：Given the function \(f(x) = x(1 - 3x^2)\), we want to find \(f^\prime(-1)\).

Step 2 ：First, we need to find the derivatives of \(u(x) = x\) and \(v(x) = 1 - 3x^2\).

Step 3 ：\(u'(x) = 1\) and \(v'(x) = -6x\).

Step 4 ：Using the product rule, \(f'(x) = u'(x)v(x) + u(x)v'(x) = 1(1 - 3x^2) + x(-6x) = 1 - 9x^2\).

Step 5 ：Finally, we evaluate the derivative at \(x = -1\): \(f'(-1) = 1 - 9(-1)^2 = 1 - 9 = -8\).

Step 6 ：\(\boxed{f^\prime(-1) = -8}\)