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

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Differentiate the equation
$$
e^y \cos(x)=4+\sin(xy)
$$
with respect to $$ x $$.

1. Differentiate the left-hand side by applying the product rule:
   $$
   \frac{d}{dx}\left(e^y \cos x\right)=\frac{d}{dx}\left(e^y\right) \cos x + e^y \frac{d}{dx}\left(\cos x\right).
   $$
   Since
   $$
   \frac{d}{dx}\left(e^y\right)=e^y \frac{dy}{dx}
   $$
   and
   $$
   \frac{d}{dx}\left(\cos x\right)=-\sin x,
   $$
   we have
   $$
   \frac{d}{dx}\left(e^y \cos x\right)=e^y \frac{dy}{dx}\cos x - e^y \sin x.
   $$

2. Differentiate the right-hand side. Since the derivative of $$4$$ is 0, differentiate $$\sin(xy)$$ using the chain rule:
   $$
   \frac{d}{dx}\left(\sin(xy)\right)=\cos(xy) \cdot \frac{d}{dx}(xy).
   $$
   The product $$xy$$ differentiates as:
   $$
   \frac{d}{dx}(xy)= y+x\frac{dy}{dx}.
   $$
   Thus,
   $$
   \frac{d}{dx}\left(\sin(xy)\right)=\cos(xy)\left(y+x\frac{dy}{dx}\right).
   $$

3. Equate the derivatives:
   $$
   e^y \frac{dy}{dx}\cos x - e^y \sin x=\cos(xy)\left(y+x\frac{dy}{dx}\right).
   $$

4. Collect the terms containing $$\frac{dy}{dx}$$:
   $$
   e^y \cos x \frac{dy}{dx} - \cos(xy) x \frac{dy}{dx} = e^y \sin x + \cos(xy) y.
   $$
   Factor out $$\frac{dy}{dx}$$:
   $$
   \frac{dy}{dx}\left(e^y \cos x - x\cos(xy)\right)= e^y \sin x + y\cos(xy).
   $$

5. Finally, solve for $$\frac{dy}{dx}$$:
   $$
   \frac{dy}{dx}=\frac{e^y \sin x + y\cos(xy)}{e^y \cos x - x\cos(xy)}.
   $$
$$ \frac{dy}{dx} = \frac{e^y \sin x + y \cos(xy)}{e^y \cos x - x \cos(xy)} $$

Find by implicit differentiation.

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