If $g(x)=\int_{x}^{7} \cos \left(t^{7}\right) d t$, then $g^{\prime}(x)=$
$-\cos \left(x^{7}\right)$
$\cos \left(7^{7}\right)-\cos \left(x^{7}\right)$
$\cos \left(x^{7}\right)$
$-\sin \left(7^{7}\right)+\sin \left(x^{7}\right)$
$-\sin \left(x^{7}\right)$

Step 1 ：Given the function \(g(x)=\int_{x}^{7} \cos \left(t^{7}\right) d t\), we are asked to find the derivative of this function.

Step 2 ：This is a problem of finding the derivative of an integral function, also known as the Fundamental Theorem of Calculus.

Step 3 ：The 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 of f from a to b is equal to F(b) - F(a).

Step 4 ：In this case, the function being integrated is \(f(t) = \cos(t^7)\), and the limits of integration are x and 7.

Step 5 ：According to the theorem, the derivative of the integral function with respect to x should be -f(x), where f(x) is the function being integrated.

Step 6 ：Therefore, the derivative of \(g(x)\) should be \(-\cos(x^7)\).

Step 7 ：\(\boxed{-\cos \left(x^{7}\right)}\) is the final answer.