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 ：Use the Fundamental Theorem of Calculus to differentiate. 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 3 ：In this case, we have a function of the form \(g(x) = \int_{u(x)}^{v(x)} f(t) dt\), where \(f(t) = e^t\), \(u(x) = \sec(x)\), and \(v(x) = \csc(x)\).

Step 4 ：The derivative of this function, according to Leibniz's rule (also known as the Fundamental Theorem of Calculus Part 1), is given by: \(g'(x) = f(v(x)) * v'(x) - f(u(x)) * u'(x)\)

Step 5 ：So, we need to find the derivatives of \(\sec(x)\) and \(\csc(x)\), which are \(\sec(x)\tan(x)\) and \(-\csc(x)\cot(x)\) respectively.

Step 6 ：Then we substitute these values into the formula.

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

Step 8 ：\(\boxed{g^\prime(x) = -e^{\csc(x)}\cot(x)\csc(x) - e^{\sec(x)}\tan(x)\sec(x)}\)