Use part I of the Fundamental Theorem of Calculus to find the derivative of $f(x)=\int_{4}^{x}\left(\frac{1}{4} t^{2}-1\right)^{5} d t$

Step 1 ：The Fundamental Theorem of Calculus Part I states that if a function f is continuous on 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 equals F(b) - F(a).

Step 2 ：In this case, we are asked to find the derivative of a function defined by an integral, which is the reverse process. The Fundamental Theorem of Calculus Part I also tells us that the derivative of the integral of a function from a to x is just the function evaluated at x.

Step 3 ：So, we need to substitute x into the function inside the integral to find the derivative of f(x).

Step 4 ：Let's substitute x into the function inside the integral: \(f = (0.25*x^2 - 1)^5\)

Step 5 ：Then, find the derivative of the function: \(derivative = 2.5*x*(0.25*x^2 - 1)^4\)

Step 6 ：Final Answer: The derivative of the function \(f(x)=\int_{4}^{x}\left(\frac{1}{4} t^{2}-1\right)^{5} d t\) is \(\boxed{2.5x\left(\frac{1}{4}x^{2}-1\right)^{4}}\)