\[ f(2)=2, f(4)=-1, f(6)=6, f(8)=4, f(10)=-3 \] Let \[ F(x)=\int_{5 x}^{x^{3}} f(t) d t \] Calculate $F^{\prime}(2)$. $-9$ 7 $-21$ 63 21

Step 1 ：We are given the function \(f(x)\) with values \(f(2)=2, f(4)=-1, f(6)=6, f(8)=4, f(10)=-3\).

Step 2 ：We are asked to calculate the derivative of the function \(F(x)=\int_{5 x}^{x^{3}} f(t) d t\) at the point \(x=2\), denoted as \(F'(2)\).

Step 3 ：By the Fundamental Theorem of Calculus, the derivative of \(F(x)\) is given by \(F'(x) = f(x^3) \cdot 3x^2 - f(5x) \cdot 5\).

Step 4 ：Substituting \(x=2\) into the formula, we get \(F'(2) = f(2^3) \cdot 3 \cdot 2^2 - f(5 \cdot 2) \cdot 5\).

Step 5 ：From the given values of the function \(f(x)\), we know that \(f(8) = 4\) and \(f(10) = -3\).

Step 6 ：Substituting these values into the formula, we get \(F'(2) = 4 \cdot 3 \cdot 2^2 - (-3) \cdot 5\).

Step 7 ：Simplifying the expression, we find that \(F'(2) = 63\).

Step 8 ：Final Answer: The derivative of the function \(F\) at the point \(x=2\), denoted as \(F'(2)\), is \(\boxed{63}\).