If $e^{x}+e^{y}=e^{x+y}$, prove that $\frac{d y}{d x}+e^{y-x}=0$
Here,
$e^{x}+e^{y}=e^{x+y} \ldots \ldots$ (i)
Differentiating both the sides using chain rule,
$\frac{d}{d x} e^{x}+\frac{d}{d x} e^{y}=\frac{d}{d x}\left(e^{x+y}\right)$
$e^{x}+e^{y} \frac{d y}{d x}=e^{x+y} \frac{d}{d x}(x+y)$
$e^{x}+e^{y} \frac{d y}{d x}=e^{x+y}\left[1+\frac{d y}{d x}\right]$
$e^{y} \frac{d y}{d x}-e^{x+y} \frac{d y}{d x}=e^{x+y}-e^{x}$
$\frac{d y}{d x}\left(e^{y}-e^{x+y}\right)=e^{x+y}-e^{x}$
$\frac{d y}{d x}=\frac{e^{x+y}-e^{x}}{e^{y}-e^{x+y}}$
$\frac{d y}{d x}=\frac{e^{x}+e^{y}-e^{x}}{e^{y}-\left(e^{x}+e^{y}\right)}$
$\frac{d y}{d x}=\frac{e^{x}+e^{y}-e^{x}}{e^{y}-e^{x}-e^{y}}$
$\frac{d y}{d x}=\frac{e^{y}}{-e^{x}}$
$\frac{d y}{d x}=-e^{y-x}$
$\frac{d y}{d x}+e^{y-x}=0$
Hence Proved.
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