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: Define the function
Let \( y = \frac{x+2}{x^2-3} \).

### Step 2: Apply the Quotient Rule
To differentiate \( y \), we will use the Quotient Rule, which states:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( u = x + 2 \) and \( v = x^2 - 3 \).

### Step 3: Differentiate \( u \) and \( v \)
Calculate \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( u = x + 2 \) implies \( \frac{du}{dx} = 1 \)
- \( v = x^2 - 3 \) implies \( \frac{dv}{dx} = 2x \)

### Step 4: Substitute into the Quotient Rule
Now substitute \( u \), \( v \), \( \frac{du}{dx} \), 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 simplify the numerator:
\[ \frac{dy}{dx} = \frac{x^2 - 3 - 2x^2 - 4x}{(x^2 - 3)^2} \]
Combine like terms:
\[ \frac{dy}{dx} = \frac{-x^2 - 4x - 3}{(x^2 - 3)^2} \]

### Step 6: Evaluate the derivative at \( x = 0 \)
Now, substitute \( x = 0 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{-(0)^2 - 4(0) - 3}{(0^2 - 3)^2} = \frac{-3}{(-3)^2} = \frac{-3}{9} = -\frac{1}{3} \]

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

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