$\begin{array}{l}G(x)=\int_{2}^{\sin x} e^{t} \ln (t) d t \\ G^{\prime}(x)=\end{array}$

Step 1 ：Given the function \(G(x)=\int_{2}^{\sin x} e^{t} \ln (t) dt\), we are asked to find its derivative \(G'(x)\).

Step 2 ：We can use the Fundamental Theorem of Calculus Part 1, which states that if a function \(f\) is continuous on the interval \([a, b]\) and \(F\) is an antiderivative of \(f\) on \([a, b]\), then \(\int_{a}^{b} f(x) dx = F(b) - F(a)\).

Step 3 ：In this case, we can consider \(e^t \ln(t)\) as \(f(t)\), and we are looking for \(F'(\sin(x))\), which is the derivative of the upper limit of the integral.

Step 4 ：Let's denote \(t = t\), \(x = x\), \(f = e^{t}\ln(t)\), and the derivative of the upper limit as \(\cos(x)\).

Step 5 ：Substituting these values into the formula, we get \(G'(x) = e^{\sin(x)} \ln(\sin(x)) \cos(x)\).

Step 6 ：Thus, the derivative of the function \(G(x)\) is \(\boxed{G'(x) = e^{\sin(x)} \ln(\sin(x)) \cos(x)}\).