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Question:

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

Solution:

Here

$e^{y}=y^{x}$

Taking log on both sides,

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

$y \log e=x \log y$

[Using $\log a^{b}=b \log a$ ]

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

Differentiating it with respect to $x$ using product rule,

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

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

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

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

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

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

$\frac{d y}{d x}=\frac{y \log y}{y-\frac{y}{\log y}}[$ Using (i) $]$

$=\frac{y \log y \times \log y}{y \log y-y}$

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

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

Hence Proved.