Product rule with exponential and logarithm derivatives

Identify the two factors: $f(x) = e^{3x}$ and $h(x) = \ln(x)$. Their derivatives are $f'(x) = 3e^{3x}$ (using the chain rule on the exponent $3x$) and $h'(x) = \frac{1}{x}$ (the standard logarithm derivative).

Apply the product rule: $g'(x) = f'(x)h(x) + f(x)h'(x) = 3e^{3x}\ln(x) + e^{3x}\cdot \frac{1}{x}$. This can be left as is or factored as $g'(x) = e^{3x}\left(3\ln(x) + \frac{1}{x}\right)$.

The term is e to the three x, over x. This term arose from multiplying the first factor, e to the three x, by the derivative of natural log of x, which is one over x. The other term, three e to the three x times natural log of x, came from differentiating the exponential factor instead.