# if ex+ey=ex+y,prove that dy/dx= -ey-x

the given equation is
${e}^{x}+{e}^{y}={e}^{x+y}\phantom{\rule{0ex}{0ex}}{e}^{x}+{e}^{y}={e}^{x}.{e}^{y}\phantom{\rule{0ex}{0ex}}\frac{{e}^{x}}{{e}^{x}.{e}^{y}}+\frac{{e}^{y}}{{e}^{x}.{e}^{y}}=1\phantom{\rule{0ex}{0ex}}{e}^{-y}+{e}^{-x}=1$
differentiating wrt x :
$-{e}^{-y}*\frac{dy}{dx}-{e}^{-x}=0\phantom{\rule{0ex}{0ex}}{e}^{-y}*\frac{dy}{dx}=-{e}^{-x}\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=-{e}^{y}.{e}^{-x}=-{e}^{y-x}$
which is the required result.
hope this helps you

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