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Question Number 62291 by rajesh4661kumar@gamil.com last updated on 19/Jun/19

Commented by maxmathsup by imad last updated on 19/Jun/19

$$\Rightarrow (e^x-1)e^y=e^x\Rightarrow e^y=\frac{e^x}{e^x-1}\Rightarrow y=ln\left(\frac{e^x}{e^x-1}\right)\Rightarrow y=x-ln(e^x-1)\Rightarrow$$$$\frac{dy}{dx}=1-\frac{e^x}{e^x-1}=\frac{e^x-1-e^x}{e^x-1}=\frac{-1}{e^x-1}.$$

Answered by tanmay last updated on 19/Jun/19

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

Commented by rajesh4661kumar@gamil.com last updated on 19/Jun/19

$$thansji$$

Commented by tanmay last updated on 19/Jun/19

$$mostwelcomesir...$$

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