Question

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

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

Answer 1

$\begin{array}{c}{e}^x+e^y=e^{x+y}\\ \frac{d}{dx}\left(e^x+e^y\right)=\frac{d}{dx}{e}^{x+y}\\ =e^x+e^y.\frac{d}{dx}=e^{x+y}\left(1+\frac{dy}{dx}\right)\\ =e^x+e^y.\frac{d}{dx}=e^{x+y}+e^{x+y}.\frac{dy}{dx}\\ =\left(e^y-e^x.e^y\right)\frac{dy}{dx}=e^x.e^y-e^x\\ =\frac{dy}{dx}=\frac{e^x\left(e^y-1\right)}{e^y\left(1-e^x\right)}\\ =\frac{dy}{dx}=e^{x-y}\left(\frac{e^y-1}{1-e^x}\right)\end{array}$