Continuity and Differentiability — Maths STD 12 Science — Question

$ \Rightarrow {x^y} = {y^x}$

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

$ \Rightarrow y\log x = x\log y$

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

$\Rightarrow y.\frac{1}{x} + \log x.\frac{{dy}}{{dx}} = x.\frac{1}{y}\frac{{dy}}{{dx}} + \log y.1$

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

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

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