1. **State the problem:** Find the derivative $\frac{dy}{dx}$ for the equation $$e^{x+y} = \cos\left(e^y\right) + \ln(xy).$$ 2. **Recall the rules:** We will use implicit differentiation since $y$ is a function of $x$. - Derivative of $e^{x+y}$ with respect to $x$ uses chain rule: $$\frac{d}{dx} e^{x+y} = e^{x+y} \cdot \frac{d}{dx}(x+y) = e^{x+y}(1 + \frac{dy}{dx}).$$ - Derivative of $\cos(e^y)$ uses chain rule: $$\frac{d}{dx} \cos(e^y) = -\sin(e^y) \cdot e^y \cdot \frac{dy}{dx}.$$ - Derivative of $\ln(xy)$ uses product rule inside the log: $$\frac{d}{dx} \ln(xy) = \frac{1}{xy} \cdot \frac{d}{dx}(xy) = \frac{1}{xy} (y + x \frac{dy}{dx}).$$ 3. **Differentiate both sides:** $$e^{x+y}(1 + \frac{dy}{dx}) = -\sin(e^y) e^y \frac{dy}{dx} + \frac{1}{xy}(y + x \frac{dy}{dx}).$$ 4. **Multiply both sides to clear fractions and group $\frac{dy}{dx}$ terms:** $$e^{x+y} + e^{x+y} \frac{dy}{dx} = -\sin(e^y) e^y \frac{dy}{dx} + \frac{y}{xy} + \frac{x}{xy} \frac{dy}{dx} = -\sin(e^y) e^y \frac{dy}{dx} + \frac{1}{x} + \frac{1}{y} \frac{dy}{dx}.$$ 5. **Rewrite grouping $\frac{dy}{dx}$ terms on one side:** $$e^{x+y} + e^{x+y} \frac{dy}{dx} = -\sin(e^y) e^y \frac{dy}{dx} + \frac{1}{x} + \frac{1}{y} \frac{dy}{dx}$$ $$e^{x+y} - \frac{1}{x} = -\sin(e^y) e^y \frac{dy}{dx} + \frac{1}{y} \frac{dy}{dx} - e^{x+y} \frac{dy}{dx}$$ 6. **Factor $\frac{dy}{dx}$ on the right:** $$e^{x+y} - \frac{1}{x} = \left(-\sin(e^y) e^y + \frac{1}{y} - e^{x+y}\right) \frac{dy}{dx}.$$ 7. **Solve for $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{e^{x+y} - \frac{1}{x}}{-\sin(e^y) e^y + \frac{1}{y} - e^{x+y}}.$$