Question
If xy = ex-y, find $\frac{\text{dy}}{\text{dx}}$
✓
Answer
The given function is xy = ex-y
Taking logarithm on both the sides, we obtain
$\log(\text{xy})=\log\big(\text{e}^{\text{x}-\text{y}})$
$\Rightarrow\log\text{x}+\log\text{y}=(\text{x}-\text{y})\log\text{e}$
$\Rightarrow\log\text{x}=\log\text{y}=(\text{x}-\text{y})\times1$
$\Rightarrow\log\text{x}=\log\text{y}=\text{x}-\text{y}$
Differentiating both sides with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\log\text{x})+\frac{\text{d}}{\text{dx}}(\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{x})-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\big(1+\frac{1}{\text{y}}\big)\frac{\text{dy}}{\text{dx}}=1-\frac{1}{\text{x}}$
$\Rightarrow\big(\frac{\text{y}+1}{\text{y}}\big)\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{\text{x}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}-1)}{\text{x}(\text{y}+1)}$
Taking logarithm on both the sides, we obtain
$\log(\text{xy})=\log\big(\text{e}^{\text{x}-\text{y}})$
$\Rightarrow\log\text{x}+\log\text{y}=(\text{x}-\text{y})\log\text{e}$
$\Rightarrow\log\text{x}=\log\text{y}=(\text{x}-\text{y})\times1$
$\Rightarrow\log\text{x}=\log\text{y}=\text{x}-\text{y}$
Differentiating both sides with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\log\text{x})+\frac{\text{d}}{\text{dx}}(\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{x})-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}}+\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=1-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\big(1+\frac{1}{\text{y}}\big)\frac{\text{dy}}{\text{dx}}=1-\frac{1}{\text{x}}$
$\Rightarrow\big(\frac{\text{y}+1}{\text{y}}\big)\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-1}{\text{x}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}-1)}{\text{x}(\text{y}+1)}$
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