To find the derivative of the function $y = 5x^2 \ln(3x)$, we’ll need to apply the product rule, which states:

$\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)$

Here, let:

• $f(x) = 5x^2$ and $g(x) = \ln(3x)$.

We’ll find $f'(x)$ and $g'(x)$ individually, then use the product rule to get $y'$.

1. Differentiate $f(x) = 5x^2$: $f'(x) = 5 \cdot 2x = 10x$

2. Differentiate $g(x) = \ln(3x)$: Using the chain rule, where $\ln(3x) = \ln(3) + \ln(x)$, we find: $g'(x) = \frac{1}{x}$

3. Apply the product rule: $y' = f'(x)g(x) + f(x)g'(x)$ Substituting the values: $y' = (10x) \ln(3x) + 5x^2 \cdot \frac{1}{x}$

4. Simplify the expression: $y' = 10x \ln(3x) + 5x$

Now, let's match this with the provided options.

The correct answer is:

[ \boxed{10