Use part I of the Fundamental Theorem of Calculus, to find the derivative of
\[
\begin{array}{l}
f(x)=\int_{-4}^{x} \sqrt{t^{3}+4^{3}} d t \\
f^{\prime}(x)=
\end{array}
\]

Step 1 ：The problem is asking for the derivative of the function \(f(x)=\int_{-4}^{x} \sqrt{t^{3}+4^{3}} dt\).

Step 2 ：We can use the Fundamental Theorem of Calculus Part 1 to solve this. This theorem 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 equal to F(b) - F(a).

Step 3 ：In this case, we are asked to find the derivative of a function defined by an integral, which is a direct application of the Fundamental Theorem of Calculus Part 1.

Step 4 ：The derivative of the function \(f(x) = \int_{-4}^{x} \sqrt{t^{3} + 4^{3}} dt\) is simply the integrand evaluated at x, which is \(\sqrt{x^{3} + 4^{3}}\).

Step 5 ：Substituting \(x = x\), we get \(f'(x) = \sqrt{x^{3} + 64}\).

Step 6 ：Final Answer: The derivative of the function is \(\boxed{\sqrt{x^{3}+64}}\).