Find \( d y / d x \) by implicit differentiation. \[ e^{y} \cos (x)=4+\sin (x y) \] \( \frac{d y}{d x}=\square \)

To find \( \frac{dy}{dx} \) by implicit differentiation for the equation \[ e^{y} \cos (x) = 4 + \sin (xy), \] we first differentiate both sides with respect to \( x \): 1. The left side: Using the product rule for \( e^y \cos(x) \): \[ \frac{d}{dx}(e^y \cos(x)) = e^y \frac{dy}{dx} \cos(x) - e^y \sin(x). \] 2. The right side: For the right side \( 4 + \sin(xy) \), we use the chain rule. The derivative of \( \sin(xy) \) requires the product rule on \( xy \): \[ \frac{d}{dx}(4 + \sin(xy)) = \cos(xy) \left( y + x \frac{dy}{dx} \right). \] Now we set these equal: \[ e^y \frac{dy}{dx} \cos(x) - e^y \sin(x) = \cos(xy) \left( y + x \frac{dy}{dx} \right). \] Rearranging gives: \[ e^y \frac{dy}{dx} \cos(x) - \cos(xy) x \frac{dy}{dx} = \cos(xy) y + e^y \sin(x). \] Factoring out \( \frac{dy}{dx} \): \[ \left( e^y \cos(x) - \cos(xy) x \right) \frac{dy}{dx} = \cos(xy) y + e^y \sin(x). \] Finally, divide by the factor in front of \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\cos(xy) y + e^y \sin(x)}{e^y \cos(x) - \cos(xy) x}. \] So, \[ \frac{d y}{d x} = \frac{\cos(xy) y + e^y \sin(x)}{e^y \cos(x) - \cos(xy) x}. \]