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If $y^{x}=e^{y-x}$ prove that $\frac{d y}{d x}=\frac{(1+\log y)^{2}}{\log y}$


Here, $y^{x}=e^{y-x}$

Taking log on both sides,

$\log y^{x}=\log e^{y-x}$

$x \log y=(y-x) \log e\left[\right.$ Since, $\left.\log (A B)=\log A+\log B ; \log a^{b}=b \log a\right]$

$x \log y=(y-x) \ldots \ldots .(i)$

Differentiating with respect to $x$ using product rule,

$\frac{d}{d x}(x \log y)=\frac{d}{d x}(y-x)$

$\left[\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{y})+\log \mathrm{y} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})\right]=\frac{\mathrm{dy}}{\mathrm{dx}}-1$

$\mathrm{x}\left(\frac{1}{\mathrm{y}}\right) \frac{\mathrm{dy}}{\mathrm{dx}}+\log \mathrm{y}(1)=\frac{\mathrm{dy}}{\mathrm{dx}}-1$

$\frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{\mathrm{y}}{(1+\log \mathrm{y}) \mathrm{y}}\right]=-(1+\log \mathrm{y})$

$\frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{1-1-\log \mathrm{y}}{(1+\log \mathrm{y})}\right]=-(1+\log \mathrm{y})$

$\frac{d y}{d x}=-\frac{(1+\log y)^{2}}{-\log y}$

$\frac{d y}{d x}=\frac{(1+\log y)^{2}}{\log y}$

Hence Proved.

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