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

$-e^{y-x}$

The correct answer is Option (3) → $-e^{y-x}$

Given:

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

Rewrite the right side:

$e^{x + y} = e^x \cdot e^y$

So the equation becomes:

$e^x + e^y = e^x e^y$

Rearranged:

$e^x e^y - e^y = e^x$

$e^y (e^x - 1) = e^x$

Therefore,

$e^y = \frac{e^x}{e^x - 1}$

Taking natural log:

$y = \ln \left( \frac{e^x}{e^x - 1} \right) = x - \ln (e^x - 1)$

Differentiate both sides w.r.t $x$:

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

Recall from above:

$e^y = \frac{e^x}{e^x - 1} \Rightarrow e^x - 1 = \frac{e^x}{e^y}$

So,

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