`(x^(2) + 3x+1) sinx`

Answer

To differentiate the expression \( y = (x^2 + 3x + 1) \sin x \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step-by-Step Solution:

1. **Identify the Functions**:
Let \( u = x^2 + 3x + 1 \) and \( v = \sin x \).

2. **Differentiate \( u \)**:
\[ \frac{du}{dx} = \frac{d}{dx}(x^2 + 3x + 1) = 2x + 3 \]

3. **Differentiate \( v \)**:
\[ \frac{dv}{dx} = \frac{d}{dx}(\sin x) = \cos x \]

4. **Apply the Product Rule**:
Using the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ \frac{dy}{dx} = (x^2 + 3x + 1) \cos x + \sin x (2x + 3) \]

5. **Simplify the Expression**:
The final expression for the derivative is:
\[ \frac{dy}{dx} = (x^2 + 3x + 1) \cos x + (2x + 3) \sin x \]

### Final Answer:
\[ \frac{dy}{dx} = (x^2 + 3x + 1) \cos x + (2x + 3) \sin x \]