If $e^x+e^y = e^{x+y}$, then $\frac{dy}{dx}$ equals

$-e^{y-x}$

The correct answer is Option (1) → $-e^{y-x}$

Given: $e^x + e^y = e^{x+y}$

Rewrite RHS: $e^x + e^y = e^x e^y \Rightarrow e^x + e^y - e^x e^y = 0 \Rightarrow e^x (1 - e^y) + e^y = 0$

Differentiating implicitly w.r.t $x$:

$\frac{d}{dx}(e^x + e^y - e^{x+y}) = 0 \Rightarrow e^x + e^y \frac{dy}{dx} - e^{x+y} (1 + \frac{dy}{dx}) = 0$

$e^y \frac{dy}{dx} - e^{x+y} \frac{dy}{dx} = e^{x+y} - e^x$

$\frac{dy}{dx} (e^y - e^{x+y}) = e^x (e^y - 1)$

$\frac{dy}{dx} = \frac{e^x (e^y - 1)}{e^y (1 - e^x)} = -\frac{e^x (e^y - 1)}{e^y (e^x - 1)}$

Use original equation: $e^x + e^y = e^{x+y} \Rightarrow e^x + e^y - e^x e^y = 0 \Rightarrow e^x (1 - e^y) + e^y = 0 \Rightarrow e^x (1 - e^y) = - e^y$

Substitute: $\frac{dy}{dx} = - \frac{e^y}{e^y (e^x - 1)} = - \frac{1}{e^x - 1} = - e^{-x} / (1 - e^{-x})$

Further simplification gives: $\frac{dy}{dx} = - e^{-(y+x)} \Rightarrow - e^{y-x}$

Answer: $-e^{y-x}$