Differentiate the following w.r.t. x :
`y= (x^2 - 3)/( x+4)`

Answer

To differentiate the function \( y = \frac{x^2 - 3}{x + 4} \) 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 the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = x^2 - 3 \) and \( v = x + 4 \).

### Step-by-Step Solution:

1. **Identify \( u \) and \( v \)**:
\[ u = x^2 - 3, \quad v = x + 4 \]

2. **Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)**:
\[ \frac{du}{dx} = \frac{d}{dx}(x^2 - 3) = 2x \]
\[ \frac{dv}{dx} = \frac{d}{dx}(x + 4) = 1 \]

3. **Apply the quotient rule**:
\[ \frac{dy}{dx} = \frac{(x + 4)(2x) - (x^2 - 3)(1)}{(x + 4)^2} \]

4. **Simplify the numerator**:
- First, calculate \( (x + 4)(2x) \):
\[ (x + 4)(2x) = 2x^2 + 8x \]
- Now, calculate \( (x^2 - 3)(1) \):
\[ (x^2 - 3)(1) = x^2 - 3 \]
- Combine these results:
\[ 2x^2 + 8x - (x^2 - 3) = 2x^2 + 8x - x^2 + 3 = x^2 + 8x + 3 \]

5. **Write the final derivative**:
\[ \frac{dy}{dx} = \frac{x^2 + 8x + 3}{(x + 4)^2} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{x^2 + 8x + 3}{(x + 4)^2} \]