Question: The derivative of \( y = \frac{3x^2 + 5}{2x - 1} \) is \( \frac{dy}{dx} = \frac{6x^2 - 6x +…

The derivative of \( y = \frac{3x^2 + 5}{2x - 1} \) is \( \frac{dy}{dx} = \frac{6x^2 - 6x + 10}{(2x - 1)^2} \).

True

False

Give your reasons

Solution

To determine if the derivative given in the problem is correct, we can find the derivative of the function using the quotient rule. The function provided is: \[ y = \frac{3x^2 + 5}{2x - 1} \] The quotient rule states: If \( y = \frac{u}{v} \), then \[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \] where \( u = 3x^2 + 5 \) and \( v = 2x - 1 \). First, calculate \( \frac{du}{dx} \): \[ \frac{du}{dx} = \frac{d}{dx}(3x^2 + 5) = 6x \] Next, calculate \( \frac{dv}{dx} \): \[ \frac{dv}{dx} = \frac{d}{dx}(2x - 1) = 2 \] Now, apply the quotient rule: \[ \frac{dy}{dx} = \frac{(2x - 1) \cdot 6x - (3x^2 + 5) \cdot 2}{(2x - 1)^2} \] Simplify the numerator: \[ = \frac{(12x^2 - 6x) - (6x^2 + 10)}{(2x - 1)^2} \] \[ = \frac{12x^2 - 6x - 6x^2 - 10}{(2x - 1)^2} \] \[ = \frac{6x^2 - 6x - 10}{(2x - 1)^2} \] The correct derivative is: \[ \frac{6x^2 - 6x - 10}{(2x - 1)^2} \] This is different from the given derivative: \[ \frac{6x^2 - 6x + 10}{(2x - 1)^2} \] Therefore, the statement is False. The correct expression in the numerator should be \(6x^2 - 6x - 10\), not \(6x^2 - 6x + 10\).