Use implicit differentiation to find \( \frac{d y}{d x} \) \[ x \sin (3 x+2 y)=y \cos x \]

To find \( \frac{d y}{d x} \) using implicit differentiation on the equation \( x \sin (3 x+2 y) = y \cos x \), we differentiate both sides with respect to \( x \). 1. Differentiate the left side using the product rule and chain rule: \[ \frac{d}{dx}(x \sin(3x + 2y)) = \sin(3x + 2y) + x \cos(3x + 2y) \cdot (3 + 2 \frac{dy}{dx}) \] 2. Differentiate the right side: \[ \frac{d}{dx}(y \cos x) = \frac{dy}{dx} \cos x - y \sin x \] Now, set the derivatives equal: \[ \sin(3x + 2y) + x \cos(3x + 2y)(3 + 2 \frac{dy}{dx}) = \frac{dy}{dx} \cos x - y \sin x \] Next, we can collect all terms involving \( \frac{dy}{dx} \) on one side: \[ x \cos(3x + 2y) \cdot 2 \frac{dy}{dx} - \frac{dy}{dx} \cos x = -\sin(3x + 2y) - y \sin x - 3x \cos(3x + 2y) \] Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left( x \cos(3x + 2y) \cdot 2 - \cos x \right) = -\sin(3x + 2y) - y \sin x - 3x \cos(3x + 2y) \] Finally, solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{-\sin(3x + 2y) - y \sin x - 3x \cos(3x + 2y)}{x \cos(3x + 2y) \cdot 2 - \cos x} \]