Find dy dx of the functions given in Exercise: xy=e^ (x-y) - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}}$ of the functions given in Exercise:
$\text{xy}=\text{e}^{(\text{x}-\text{y})}$

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Answer

Given: $\text{xy}=\text{e}^{\text{x}-\text{y}}\ \Rightarrow\ \log\text{xy}\ \log=\text{e}^{\text{x}-\text{y}}$
$\Rightarrow\ \log\text{x}+\log\text{y}=(\text{x}-\text{y})\log\text{e}\ \Rightarrow\ \log\text{x}+\log\text{y}=(\text{x}-\text{y})\ \ [\because\log\text{e}=1]$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{x}+\frac{\text{d}}{\text{dx}}\log\text{y}=\frac{\text{d}}{\text{dx}}(\text{x}-\text{y})\ \Rightarrow\ \frac{1}{\text{x}}+\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}+\frac{\text{dy}}{\text{dx}}=1-\frac{1}{\text{x}}\ \Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}+1\Big)=\frac{\text{x}-1}{\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big(\frac{1+\text{y}}{\text{y}}\Big)=\frac{\text{x}-1}{\text{x}}\ \Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}-1)}{\text{x}(1+\text{y})}$

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