Differentiate `(x+2)/(x^(2)-3)` and find the value of the derivative at x = 0.

Answer

To differentiate the function \( y = \frac{x + 2}{x^2 - 3} \) and find the value of the derivative at \( x = 0 \), we will follow these steps:

### Step 1: Identify the function
We start with the function:
\[ y = \frac{x + 2}{x^2 - 3} \]

### Step 2: Apply the Quotient Rule
The quotient rule states that if \( y = \frac{u}{v} \), then the derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, \( u = x + 2 \) and \( v = x^2 - 3 \).

### Step 3: Differentiate \( u \) and \( v \)
Now we differentiate \( u \) and \( v \):
- \( \frac{du}{dx} = 1 \) (since the derivative of \( x + 2 \) is 1)
- \( \frac{dv}{dx} = 2x \) (since the derivative of \( x^2 - 3 \) is \( 2x \))

### Step 4: Substitute into the Quotient Rule
Now, substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule:
\[ \frac{dy}{dx} = \frac{(x^2 - 3)(1) - (x + 2)(2x)}{(x^2 - 3)^2} \]

### Step 5: Simplify the expression
Now we simplify the numerator:
\[ = \frac{x^2 - 3 - (2x^2 + 4x)}{(x^2 - 3)^2} \]
\[ = \frac{x^2 - 3 - 2x^2 - 4x}{(x^2 - 3)^2} \]
\[ = \frac{-x^2 - 4x - 3}{(x^2 - 3)^2} \]

### Step 6: Find the value of the derivative at \( x = 0 \)
Now we need to find \( \frac{dy}{dx} \) at \( x = 0 \):
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{-(0)^2 - 4(0) - 3}{(0^2 - 3)^2} \]
\[ = \frac{-3}{(-3)^2} \]
\[ = \frac{-3}{9} \]
\[ = -\frac{1}{3} \]

### Final Answer
Thus, the value of the derivative at \( x = 0 \) is:
\[ \frac{dy}{dx} \bigg|_{x=0} = -\frac{1}{3} \]