Question
If $y^x = e^{y-x}, $Prove that $\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
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Answer
Here,
$y^x = e^{y-x}$
Taking log on both the sides,
$\log\text{y}=\log\text{e}^{(\text{y}-\text{x})}$
$\big[\text{Since},\log\text{a}^{\text{b}}=\text{b}\log\text{a and}\log_\text{e}\text{e}=1\big]$
$\text{x}\log\text{y}=(\text{y}-\text{x})\log\text{e}$
$\text{x}\log\text{y}=\text{y}-\text{x}\ .....(\text{i})$
Differentiating it with respect to x using product rule,
$\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{y}-\text{x})$
$\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]=\frac{\text{dy}}{\text{dx}}-1$
$\text{x}\Big(\frac{\text{x}}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)=\frac{\text{dy}}{\text{dx}}-1$
$\frac{\text{dx}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}-1\Big)=-1-\log\text{y}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{y}}{(1+\log\text{y})\text{y}}\Big)=-(1+\log\text{y})$
$\Big[\text{Since, from equation (i), x}=\frac{\text{y}}{(1+\log\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{1-1-\log\text{y}}{(1+\log\text{y})}\Big]=-(1+\log\text{y})$
$\frac{\text{dy}}{\text{dx}}=-\frac{(1+\log\text{y})^2}{-\log\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
$y^x = e^{y-x}$
Taking log on both the sides,
$\log\text{y}=\log\text{e}^{(\text{y}-\text{x})}$
$\big[\text{Since},\log\text{a}^{\text{b}}=\text{b}\log\text{a and}\log_\text{e}\text{e}=1\big]$
$\text{x}\log\text{y}=(\text{y}-\text{x})\log\text{e}$
$\text{x}\log\text{y}=\text{y}-\text{x}\ .....(\text{i})$
Differentiating it with respect to x using product rule,
$\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{y}-\text{x})$
$\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]=\frac{\text{dy}}{\text{dx}}-1$
$\text{x}\Big(\frac{\text{x}}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)=\frac{\text{dy}}{\text{dx}}-1$
$\frac{\text{dx}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}-1\Big)=-1-\log\text{y}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{y}}{(1+\log\text{y})\text{y}}\Big)=-(1+\log\text{y})$
$\Big[\text{Since, from equation (i), x}=\frac{\text{y}}{(1+\log\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{1-1-\log\text{y}}{(1+\log\text{y})}\Big]=-(1+\log\text{y})$
$\frac{\text{dy}}{\text{dx}}=-\frac{(1+\log\text{y})^2}{-\log\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
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