Question

E^x/y=x-y find dy/dx by implicit differentiation

Answer 1

`\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} \\\\\\`


`\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{(y-e^x)/(y)}{(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}`