$\operatorname{Let} g(x)=\int_{\sec (x)}^{\csc (x)} e^{t} d t$
Use the Fundamental Theorem of Calculus to differentiate.
\[
g^{\prime}(x)=
\]

Step 1 ：Let \( g(x)=\int_{\sec (x)}^{\csc (x)} e^{t} d t \)

Step 2 ：We are asked to find the derivative of the function \( g(x) \), which is defined as the definite integral of \( e^t \) from \( \sec(x) \) to \( \csc(x) \).

Step 3 ：The Fundamental Theorem of Calculus 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 ：The derivative of a definite integral with respect to its upper limit is just the integrand evaluated at the upper limit, and the derivative with respect to its lower limit is the negative of the integrand evaluated at the lower limit.

Step 5 ：So, we need to find the derivative of \( \sec(x) \) and \( \csc(x) \), and then substitute these values into \( e^t \).

Step 6 ：The derivative of the function \( g(x) \) is given by the expression \( e^{\csc(x)}\cot(x)^{2}\csc(x)^{2} - e^{\sec(x)}\tan(x)^{2}\sec(x)^{2} \). This is the result of applying the Fundamental Theorem of Calculus and the chain rule of differentiation.

Step 7 ：Final Answer: \( g^{\prime}(x)=\boxed{e^{\csc(x)}\cot(x)^{2}\csc(x)^{2} - e^{\sec(x)}\tan(x)^{2}\sec(x)^{2}} \)