Question
Find $\frac{{dy}}{{dx}}$, of the function $y^x = x^y$

Answer

Given:$y^x = x^y$
$\Rightarrow {x^y} = {y^x}$
$\Rightarrow \log {x^y} = \log {y^x}$
$ \Rightarrow y\log x = x\log y$
$\Rightarrow \frac{d}{{dx}}\left( {y\log x} \right) = \frac{d}{{dx}}\left( {x\log y} \right)$
$\Rightarrow y.\frac{1}{x} + \log x.\frac{{dy}}{{dx}} = x.\frac{1}{y}\frac{{dy}}{{dx}} + \log y.1$
$\Rightarrow \left( {\log x - \frac{x}{y}} \right)\frac{{dy}}{{dx}} = \log y - \frac{y}{x}$
$\Rightarrow \left( {\frac{{y\log x - x}}{y}} \right)\frac{{dy}}{{dx}} = \frac{{x\log y - y}}{x}$
$\Rightarrow \frac{{dy}}{{dx}} = \frac{{y\left( {x\log y - y} \right)}}{{x\left( {y\log x - x} \right)}}$

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