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If $e^{x}+e^{y}=e^{x+y}$, prove that $\frac{d y}{d x}=-\frac{e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}$ or, $\frac{d y}{d x}, e^{y-x}=0$


We are given with an equation $\mathrm{e}^{x}+\mathrm{e}^{y}=\mathrm{e}^{x+y}$, we have to prove that $\frac{d y}{d x}=\frac{-e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}$ by using the given

equation we will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to $x$, we get,

$e^{x}+e^{y} \frac{d y}{d x}=e^{(x+y)}\left[1+\frac{d y}{d x}\right]$

$\frac{d y}{d x}\left[e^{y}-e^{x+y}\right]=e^{x+y}-e^{x}$

$\frac{d y}{d x}=\frac{e^{x+y}-e^{x}}{e^{y}-e^{x+y}}$

$\frac{d y}{d x}=\frac{-e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}$

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