Question

If $x^y=y^x$, then find $\frac{dy}{dx}$.

Answer 1

$x^y=y^x.$

Taking $log$ on both sides

$y\log x=x\log y.$

Differentiating both the sides by $uv$ rule

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

$\Rightarrow \frac{y}{x}-logy=\frac{x}{y}.\frac{dy}{dx}-logx\frac{dy}{dx}$

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

$\Rightarrow \frac{dy}{dx}=\frac{\frac{y-xlogy}{x}}{\frac{x-ylogx}{y}}$

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