Use the quotient rule to find the derivative of the following. \( y=\frac{x^{2}-3 x+7}{x^{2}+8} \)

To find the derivative of the function \( y = \frac{x^2 - 3x + 7}{x^2 + 8} \) using the quotient rule, we start by recalling the quotient rule formula: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \] where \( u = x^2 - 3x + 7 \) and \( v = x^2 + 8 \). First, we need to find the derivatives \( u' \) and \( v' \): 1. **Finding \( u' \)**: \[ u = x^2 - 3x + 7 \] \[ u' = \frac{d}{dx}(x^2) - \frac{d}{dx}(3x) + \frac{d}{dx}(7) = 2x - 3 + 0 = 2x - 3 \] 2. **Finding \( v' \)**: \[ v = x^2 + 8 \] \[ v' = \frac{d}{dx}(x^2) + \frac{d}{dx}(8) = 2x + 0 = 2x \] Now we can apply the quotient rule: \[ y' = \frac{(2x - 3)(x^2 + 8) - (x^2 - 3x + 7)(2x)}{(x^2 + 8)^2} \] Next, we will simplify the numerator: 1. **Expanding \( (2x - 3)(x^2 + 8) \)**: \[ (2x - 3)(x^2 + 8) = 2x^3 + 16x - 3x^2 - 24 = 2x^3 - 3x^2 + 16x - 24 \] 2. **Expanding \( (x^2 - 3x + 7)(2x) \)**: \[ (x^2 - 3x + 7)(2x) = 2x^3 - 6x^2 + 14x \] Now, substituting these expansions back into the derivative: \[ y' = \frac{(2x^3 - 3x^2 + 16x - 24) - (2x^3 - 6x^2 + 14x)}{(x^2 + 8)^2} \] Now, simplify the numerator: \[ y' = \frac{2x^3 - 3x^2 + 16x - 24 - 2x^3 + 6x^2 - 14x}{(x^2 + 8)^2} \] Combine like terms: \[ y' = \frac{(-3x^2 + 6x^2) + (16x - 14x) - 24}{(x^2 + 8)^2} \] \[ y' = \frac{3x^2 + 2x - 24}{(x^2 + 8)^2} \] Thus, the derivative of the function is: \[ \boxed{\frac{3x^2 + 2x - 24}{(x^2 + 8)^2}} \]