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Question

# If $xy={e}^{x-y},\mathrm{find}\frac{dy}{dx}$.

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Solution

## $\mathrm{We}\mathrm{have},xy={e}^{\left(x-y\right)}$ Taking log on both sides, $\mathrm{log}\left(xy\right)=\mathrm{log}\left({e}^{x-y}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{log}x+\mathrm{log}y=\left(x-y\right)\mathrm{log}e\phantom{\rule{0ex}{0ex}}⇒\mathrm{log}x+\mathrm{log}y=\left(x-y\right)×1\phantom{\rule{0ex}{0ex}}⇒\mathrm{log}x+\mathrm{log}y=x-y$ $⇒\frac{d}{dx}\left(\mathrm{log}x\right)+\frac{d}{dx}\left(\mathrm{log}y\right)=\frac{d}{dx}\left(x\right)-\frac{dy}{dx}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}\phantom{\rule{0ex}{0ex}}⇒\left(1+\frac{1}{y}\right)\frac{dy}{dx}=1-\frac{1}{x}\phantom{\rule{0ex}{0ex}}⇒\left(\frac{y+1}{y}\right)\frac{dy}{dx}=\frac{x-1}{x}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{y\left(x-1\right)}{x\left(y+1\right)}$

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