Find the derivative of the following function. \[ y=\frac{\left(2 x^{2}+4\right)(3 x+1)}{4 x-3} \]

To find the derivative of \( y = \frac{(2x^2 + 4)(3x + 1)}{4x - 3} \), we can use the quotient rule, which is given by: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \] Here, let \( u = (2x^2 + 4)(3x + 1) \) and \( v = 4x - 3 \). First, we need to compute \( u' \) and \( v' \). 1. **Finding \( v' \)**: \[ v' = \frac{d}{dx}(4x - 3) = 4 \] 2. **Finding \( u' \)**: For \( u = (2x^2 + 4)(3x + 1) \), we will use the product rule: Let \( a = 2x^2 + 4 \) and \( b = 3x + 1 \). - Find \( a' \): \[ a' = \frac{d}{dx}(2x^2 + 4) = 4x \] - Find \( b' \): \[ b' = \frac{d}{dx}(3x + 1) = 3 \] - Apply the product rule: \[ u' = a'b + ab' = (4x)(3x + 1) + (2x^2 + 4)(3) \] Expanding this: \[ u' = 12x^2 + 4x + 6x^2 + 12 = 18x^2 + 4x + 12 \] 3. **Putting it all together**: Now we can use the quotient rule: \[ y' = \frac{(u'v - uv')}{v^2} \] Substitute \( u' \), \( v \), \( u \), and \( v' \): \( u' = 18x^2 + 4x + 12 \) \( v = 4x - 3 \) \( u = (2x^2 + 4)(3x + 1) \) \( v' = 4 \) Therefore: \[ y' = \frac{(18x^2 + 4x + 12)(4x - 3) - (2x^2 + 4)(3x + 1)(4)}{(4x - 3)^2} \] 4. **Simplifying**: To simplify the expression, you would expand both parts and then combine like terms. The final answer (derivative) is: \[ y' = \frac{(18x^2 + 4x + 12)(4x - 3) - 4(2x^2 + 4)(3x + 1)}{(4x - 3)^2} \] This will give the derivative of the function \( y \).