`((x^(2) + 3x-1)/(x+2))`

Answer

To differentiate the function \( f(x) = \frac{x^2 + 3x - 1}{x + 2} \) with respect to \( x \), we will use the quotient rule. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then its derivative is given by:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]

### Step-by-Step Solution:

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

2. **Differentiate \( u \) and \( v \)**:
- \( \frac{du}{dx} = 2x + 3 \)
- \( \frac{dv}{dx} = 1 \)

3. **Apply the Quotient Rule**:
Using the quotient rule formula:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{(x + 2)(2x + 3) - (x^2 + 3x - 1)(1)}{(x + 2)^2} \]

4. **Simplify the numerator**:
- First, expand \( (x + 2)(2x + 3) \):
\[ (x + 2)(2x + 3) = 2x^2 + 3x + 4x + 6 = 2x^2 + 7x + 6 \]
- Now, subtract \( (x^2 + 3x - 1) \):
\[ 2x^2 + 7x + 6 - (x^2 + 3x - 1) = 2x^2 + 7x + 6 - x^2 - 3x + 1 = x^2 + 4x + 7 \]

5. **Write the final derivative**:
\[ f'(x) = \frac{x^2 + 4x + 7}{(x + 2)^2} \]

### Final Answer:
The derivative of \( f(x) = \frac{x^2 + 3x - 1}{x + 2} \) is:
\[ f'(x) = \frac{x^2 + 4x + 7}{(x + 2)^2} \]