Find \( \mathrm{d} y / \mathrm{d} x \) \( x \sin y=3 x y+y \) Type answer \( y^{\prime}= \) (numerator)/(denominator) A,

To find \( \mathrm{d} y / \mathrm{d} x \) for the equation \( x \sin y = 3xy + y \), we'll use implicit differentiation. 1. Differentiate both sides with respect to \( x \): - Left side: \( \frac{d}{dx}(x \sin y) = \sin y + x \cos y \frac{dy}{dx} \) - Right side: \( \frac{d}{dx}(3xy + y) = 3y + 3x \frac{dy}{dx} + \frac{dy}{dx} \) 2. This gives: \[ \sin y + x \cos y \frac{dy}{dx} = 3y + (3x + 1) \frac{dy}{dx} \] 3. Rearranging terms, we get: \[ x \cos y \frac{dy}{dx} - (3x + 1) \frac{dy}{dx} = 3y - \sin y \] 4. Factor out \( \frac{dy}{dx} \): \[ \left( x \cos y - (3x + 1) \right) \frac{dy}{dx} = 3y - \sin y \] 5. Solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{3y - \sin y}{x \cos y - (3x + 1)} \] Thus, the final result is: \[ y^{\prime} = \frac{3y - \sin y}{x \cos y - (3x + 1)} \]