Answer is: option3

0

Solution:

We differentiate both sides of the first derivative equation with respect to \( x \): \[ \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{-3x^2 - y}{x} \right) \] Using the quotient rule: \[ \frac{d^2y}{dx^2} = \frac{x(-6x - \frac{dy}{dx}) - (-3x^2 - y)(1)}{x^2} \] \[ = \frac{-6x^2 - x \frac{dy}{dx} + 3x^2 + y}{x^2} \] \[ = \frac{-3x^2 - x \frac{dy}{dx} + y}{x^2} \] Find \( y \) when \( x = 2 \)

From the equation of the curve:

\[ 2^3 + 2y = 8 \] \[ 8 + 2y = 8 \] \[ 2y = 0 \] \[ y = 0 \] Find \( \frac{dy}{dx} \) at \( x = 2 \) \[ \frac{dy}{dx} = \frac{-3(2)^2 - 0}{2} \] \[ = \frac{-12}{2} = -6 \] Compute \( \frac{d^2y}{dx^2} \) at \( x = 2 \) \[ \frac{d^2y}{dx^2} = \frac{-3(2)^2 - (2)(-6) + 0}{2^2} \] \[ = \frac{-12 + 12}{4} \] \[ = \frac{0}{4} = 0 \] Final Answer: 0