Question

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

Answer 1

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

differentiating above with respect to $x$ we get,

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

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

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

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

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

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

Also $1+e^{y-x}=e^y$

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

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

Also $e^{x-y}+1=e^x$

$\therefore e^{x-y}+1=e^x$

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