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If $e^x+e^y=e^{x+y}$, then $\frac{d y}{d x}=$
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The correct answer is:
$-e^{y-x}$
$e^x+e^y=e^{x+y} \Rightarrow e^x+e^y \cdot \frac{d y}{d x}=e^{x+y}\left(1+\frac{d y}{d x}\right)$
$\Rightarrow \frac{d y}{d x}=\frac{e^x-e^{x+y}}{e^{x+y}-e^y}=\frac{e^x-e^x-e^y}{e^x+e^y-e^y}=-\frac{e^y}{e^x}=-e^{y-x}$
$\Rightarrow \frac{d y}{d x}=\frac{e^x-e^{x+y}}{e^{x+y}-e^y}=\frac{e^x-e^x-e^y}{e^x+e^y-e^y}=-\frac{e^y}{e^x}=-e^{y-x}$
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