Q.If $e^x + e^y = e^{(x + y)}$, then show that $\frac{dy}{dx} = -e^{y - x}$.

Differentiate $e^x + e^y = e^{x+y}$ implicitly, collect the $\frac{dy}{dx}$ terms, and substitute $e^{x+y} = e^x + e^y$ (the given relation) to get $\frac{dy}{dx} = -e^{y-x}$.

Given $e^x + e^y = e^{x+y}$. Differentiate both sides with respect to $x$, treating $y$ as a function of $x$:

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

Expand the right side:

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

Collect the $\frac{dy}{dx}$ terms on one side:

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

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

…