Respuesta :

[tex]\bf \cfrac{e^x}{y}=x-y\implies e^xy^{-1}=x-y \\\\\\ e^x\cdot y^{-1}+e^x\cdot -1y^{-2}\cfrac{dy}{dx}=1-\cfrac{dy}{dx} \\\\\\ \cfrac{e^x}{y}-\cfrac{e^x}{y^2}\cdot \cfrac{dy}{dx}=1-\cfrac{dy}{dx} \\\\\\ \cfrac{dy}{dx}-\cfrac{e^x}{y^2}\cdot \cfrac{dy}{dx}=1-\cfrac{e^x}{y}\impliedby \textit{common factor} \\\\\\[/tex]

[tex]\bf \cfrac{dy}{dx}\left( 1-\cfrac{e^x}{y^2} \right)=1-\cfrac{e^x}{y}\implies \cfrac{dy}{dx}\left(\cfrac{y-e^x}{y^2} \right)=\cfrac{y-e^x}{y} \\\\\\ \cfrac{dy}{dx}=\cfrac{\frac{y-e^x}{y}}{\frac{y-e^x}{y^2} }\implies \cfrac{dy}{dx}=\cfrac{y-e^x}{y}\cdot \cfrac{y^2}{y-e^x}\implies \cfrac{dy}{dx}=\cfrac{y(y-e^x)}{y-e^x}[/tex]