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

Answer

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Answer

$\boxed{f^\prime(-1) = -8}$

Steps

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 ：$\boxed{f^\prime(-1) = -8}$