Question
If $\text{x}-\text{e}^{\frac{\text{x}}{\text{y}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
✓
Answer
$\text{x}-\text{e}^{\frac{\text{x}}{\text{y}}}$
Taking logarithm on both sides, we get
$\log\text{x}=\frac{\text{x}}{\text{y}}$
$\Rightarrow\text{y}\log\text{x}=\text{x}$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=1$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}=1-\frac{\text{y}}{\text{x}}$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}}$
$\Rightarrow \text{x}\log\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}-\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
Taking logarithm on both sides, we get
$\log\text{x}=\frac{\text{x}}{\text{y}}$
$\Rightarrow\text{y}\log\text{x}=\text{x}$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=1$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}=1-\frac{\text{y}}{\text{x}}$
$\Rightarrow\log\text{x}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}}$
$\Rightarrow \text{x}\log\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}-\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}$
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