Question

Find $\frac{dy}{dx}$ of $y^x=x^y$

Answer 1

The given function is $y^x=x^y$
Taking logarithm on both the sides, we obtain
$x\log y=y\log x$
Differentiating both sides with respect to $x$, we obtain
$\log y.\frac{d}{dx}\left(x\right)+x.\frac{d}{dx}\left(\log y\right)=\log x.\frac{d}{dx}\left(y\right)+y.\frac{d}{dx}\left(\log x\right)$
$\Rightarrow \log y.1+x.\frac{1}{y}.\frac{dy}{dx}=\log x.\frac{dy}{dx}+y.\frac{1}{x}$
$\Rightarrow \log y+\frac{x}{y}\frac{dy}{dx}=\log x.\frac{dy}{dx}+\frac{y}{x}$
$\Rightarrow (\frac{x}{y}-\log x)\frac{dy}{dx}=\frac{y}{x}-\log y$
$\Rightarrow \left(\frac{x-y\log x}{y}\right)\frac{dy}{dx}=\frac{y-x\log y}{x}$
$\therefore \frac{dy}{dx}=\frac{y}{x}\left(\frac{y-x\log y}{x-y\log x}\right)$