Find the derivative of `f (x) = (2x + 1)/(x + 3)`

Answer

To find the derivative of the function \( f(x) = \frac{2x + 1}{x + 3} \), we will use the quotient rule. The quotient rule states that if you have a function in the form of \( \frac{u}{v} \), where \( u \) and \( v \) are both functions of \( x \), then the derivative \( f'(x) \) is given by:

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

### Step-by-Step Solution:

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

2. **Find the derivatives \( u' \) and \( v' \)**:
- Differentiate \( u \):
\[ u' = \frac{d}{dx}(2x + 1) = 2 \]
- Differentiate \( v \):
\[ v' = \frac{d}{dx}(x + 3) = 1 \]

3. **Apply the quotient rule**:
- Substitute \( u \), \( v \), \( u' \), and \( v' \) into the quotient rule formula:
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
- Plugging in the values:
\[ f'(x) = \frac{(2)(x + 3) - (2x + 1)(1)}{(x + 3)^2} \]

4. **Simplify the numerator**:
- Expand the terms in the numerator:
\[ f'(x) = \frac{2x + 6 - (2x + 1)}{(x + 3)^2} \]
- Combine like terms:
\[ f'(x) = \frac{2x + 6 - 2x - 1}{(x + 3)^2} = \frac{5}{(x + 3)^2} \]

5. **Final result**:
- Therefore, the derivative of the function is:
\[ f'(x) = \frac{5}{(x + 3)^2} \]