Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \[ \begin{array}{r} h(x)=\int_{1}^{e^{x}} 6 \ln (t) d t \\ h^{\prime}(x)=\square_{x} \end{array} \]

Step 1 ：The Fundamental Theorem of Calculus Part 1 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 F(b) - F(a).

Step 2 ：In this case, we are asked to find the derivative of the function h(x), which is defined as the definite integral from 1 to e^x of 6ln(t) dt. According to the Fundamental Theorem of Calculus Part 1, the derivative of this function with respect to x is simply the integrand evaluated at the upper limit of integration, e^x, times the derivative of the upper limit with respect to x.

Step 3 ：So, we need to substitute e^x into the integrand 6ln(t) and then multiply by the derivative of e^x with respect to x.

Step 4 ：The derivative of the function h(x) is 6. This is the result of substituting e^x into the integrand 6ln(t) and then multiplying by the derivative of e^x with respect to x.

Step 5 ：Final Answer: The derivative of the function h(x) is \(\boxed{6}\).