Use Part 2 of the Fundamental Theorem of Calculus to find each of the following when $f(x)=\int_{4}^{x} t^{5} d t$. \[ \begin{array}{l} f^{\prime}(x)= \\ f^{\prime}(-2)= \end{array} \]

Step 1 ：The Fundamental Theorem of Calculus Part 2 states that if a function f is continuous over the interval [a, b] and F is an antiderivative of f on [a, b], then the definite integral from a to b of f(x) dx is F(b) - F(a).

Step 2 ：In this case, we are given that f(x) is the integral from 4 to x of t^5 dt. This means that the derivative of f(x), f'(x), is simply the integrand, t^5, evaluated at x.

Step 3 ：So, f'(x) = x^5.

Step 4 ：To find f'(-2), we simply substitute -2 for x in the expression for f'(x).

Step 5 ：After substituting, we find that f'(-2) = -32.

Step 6 ：So, the final answers are \(f^{\prime}(x) = \boxed{x^{5}}\) and \(f^{\prime}(-2) = \boxed{-32}\).