`e^(2x) sin 3x`

Answer

To differentiate the function \( y = e^{2x} \sin(3x) \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their product is given by:

\[ \frac{d}{dx}(u \cdot v) = u'v + uv' \]

### Step-by-step solution:

1. **Identify the functions**:
Let \( u = e^{2x} \) and \( v = \sin(3x) \).

2. **Differentiate \( u \)**:
To differentiate \( u = e^{2x} \), we apply the chain rule:
\[ u' = \frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x) = e^{2x} \cdot 2 = 2e^{2x} \]

3. **Differentiate \( v \)**:
To differentiate \( v = \sin(3x) \), we again use the chain rule:
\[ v' = \frac{d}{dx}(\sin(3x)) = \cos(3x) \cdot \frac{d}{dx}(3x) = \cos(3x) \cdot 3 = 3\cos(3x) \]

4. **Apply the product rule**:
Now, we can apply the product rule:
\[ \frac{dy}{dx} = u'v + uv' = (2e^{2x}) \sin(3x) + (e^{2x})(3\cos(3x)) \]

5. **Combine the terms**:
Factor out \( e^{2x} \):
\[ \frac{dy}{dx} = e^{2x}(2\sin(3x) + 3\cos(3x)) \]

### Final answer:
Thus, the derivative of \( y = e^{2x} \sin(3x) \) is:
\[ \frac{dy}{dx} = e^{2x}(2\sin(3x) + 3\cos(3x)) \]