If e^y= y^x, prove that dy dx = ( )^2 -1

Question

If $e^y= y^x,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{(\log\text{y})^2}{\log\text{y}-1}$

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Answer

We have, $e^y = y^x$
Taking log on both sides,
$\log\text{e}^{\text{y}}=\log\text{y}^\text{x}$
$\Rightarrow\text{y}\log\text{e}=\text{x}\log\text{y}$
$\Rightarrow\text{y}=\text{x}\log\text{y}\ .....(\text{i})$
Differentiating with respect to $x,$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big(1-\frac{\text{x}}{\text{y}}\Big)=\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big(\frac{\text{y}-\text{x}}{\text{y}}\big)=\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}\log\text{y}}{\text{y}-\text{x}}$
$\Rightarrow\frac{\text{y}\log\text{y}}{\Big(\text{y}-\frac{\text{y}}{\log\text{y}}\Big)}$
$[$Using equation $(i)]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}\log\text{y}(\log\text{y})}{\text{y}\log\text{y}-\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\log\text{y})^2}{\text{y}(\log\text{y}-1)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\log\text{y})^2}{(\log\text{y}-1)}$

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