If `y = sin x l n(3x)` then `(dy)/(dx)` will be :

Options

A. `(3 cos x)/(x)`
B. `(sin x)/(x)+(cos x)l n(3x)`
C. `sin (x^(2))+cos x l n(3x)`
D. `3 x sin x+3 xos x l n(x)`

Explanation

To find the derivative of the function \( y = \sin x \ln(3x) \), we will use the product rule of differentiation. The product rule states that if you have two functions multiplied together, \( u \) and \( v \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = \frac{du}{dx} v + u \frac{dv}{dx} \]

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( u = \sin x \) and \( v = \ln(3x) \).

2. **Differentiate \( u \)**:
The derivative of \( u \) is:
\[ \frac{du}{dx} = \cos x \]

3. **Differentiate \( v \)**:
To differentiate \( v = \ln(3x) \), we use the chain rule. The derivative of \( \ln(x) \) is \( \frac{1}{x} \), so:
\[ \frac{dv}{dx} = \frac{1}{3x} \cdot 3 = \frac{1}{x} \]

4. **Apply the product rule**:
Now, substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the product rule formula:
\[ \frac{dy}{dx} = \frac{du}{dx} v + u \frac{dv}{dx} \]
This becomes:
\[ \frac{dy}{dx} = \cos x \cdot \ln(3x) + \sin x \cdot \frac{1}{x} \]

5. **Simplify the expression**:
Thus, we have:
\[ \frac{dy}{dx} = \cos x \ln(3x) + \frac{\sin x}{x} \]

### Final Answer:
The derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \cos x \ln(3x) + \frac{\sin x}{x} \]