1. Compute the derivative of \( g(x) = 5 + e^{2x} + \ln{6} \):
g'(x) = \frac{d}{dx} \left( 5 + e^{2x} + \ln{6} \right)
2. Use the constant rule and the chain rule for differentiation:
\frac{d}{dx}(5) = 0
\frac{d}{dx}(e^{2x}) = 2e^{2x}
\frac{d}{dx}(\ln{6}) = 0
3. Combine the results:
g'(x) = 0 + 2e^{2x} + 0
g'(x) = 2e^{2x}
4. Compare \( g'(x) \) with \( h(x) \):
2e^{2x} \neq 3e^{2x} + \ln{4}
Since \( g'(x) \neq h(x) \), the function \( g(x) = 5 + e^{2x} + \ln{6} \) is not a primitive function of \( h(x) = 3e^{2x} + \ln{4} \).
Therefore, the answer is:
\boxed{\text{No, } g(x) = 5 + e^{2x} + \ln{6} \text{ is not a primitive function of } h(x) = 3e^{2x} + \ln{4}}