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

To find \( \frac{d y}{d x} \) by implicit differentiation, we start by differentiating both sides of the equation with respect to \( x \): Given: \[ e^{y} \cos (x)=4+\sin (x y) \] Differentiate the left side: Using the product rule on \( e^y \cos(x) \): \[ \frac{d}{dx}(e^{y}\cos(x)) = e^{y} \frac{dy}{dx} \cos(x) - e^{y} \sin(x) \] Differentiate the right side: Using the chain rule on \( \sin(xy) \): \[ \frac{d}{dx}(4+\sin(xy)) = \cos(xy) \left( y + x \frac{dy}{dx} \right) \] Now, we equate both derivatives: \[ e^{y} \left( \frac{dy}{dx} \cos(x) - \sin(x) \right) = \cos(xy) \left( y + x \frac{dy}{dx} \right) \] Now, rearranging to isolate \( \frac{dy}{dx} \): \[ e^{y} \frac{dy}{dx} \cos(x) - \cos(xy) x \frac{dy}{dx} = \cos(xy) y + e^{y} \sin(x) \] \[ \frac{dy}{dx} \left( e^{y} \cos(x) - \cos(xy) x \right) = \cos(xy) y + e^{y} \sin(x) \] Thus, we can solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\cos(xy) y + e^{y} \sin(x)}{e^{y} \cos(x) - \cos(xy) x} \] Finally, the answer is: \[ \frac{d y}{d x} = \frac{\cos(xy) y + e^{y} \sin(x)}{e^{y} \cos(x) - \cos(xy) x} \]