Question
If $e ^{ x }+ e ^{ y }= e ^{ x + y }$, prove that $\frac{d y}{d x}+ e ^{ y - x }=0$.

Answer

Given $e^{ x }+ e ^{ y }= e ^{ x + y }$
On dividing Eq(i) by $e ^{ x + y }$, we get,
$e^{-y}+e^{-x}=1 \ldots \text { (ii) }$
Therefore, on differentiating both sides of Eq(ii) w.r.t x, we get,
$e^{-y} \cdot\left(\frac{-d y}{d x}\right)+e^{-x}(-1)=0$
$\Rightarrow-e^{-y} \frac{d y}{d x}+e^{-x}(-1)=0$
$\Rightarrow-e^{-y} \frac{d y}{d x}=e^{-x}$
$\Rightarrow \frac{d y}{d x}=-\frac{e^{-x}}{e^{-y}}$
$\Rightarrow \frac{d y}{d x}=-e^{(y-x)}$
$\therefore \frac{d y}{d x}+e^{(y-x)}=0$
Hence Proved.

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