Question

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

Answer 1

The correct answer is $∴\frac{dy}{dx}=\frac{y}{x}(\frac{y-xlogy}{x-ylogx})$
The given function is $y^x=x^y$
Taking logarithm on both the sides,we obtain
$x\,logy=y\,logx$
Differentiating both sides with respect to $x$,we obtain
$log\,y.\frac{d}{dx}(x)+x.\frac{d}{dx}(logy)=logx.\frac{d}{dy}(y)+y.\frac{d}{dx}(logx)$
$⇒logy.1+x.\frac{1}{y}.\frac{dy}{dx}=logx.\frac{dy}{dx}+y.\frac{1}{x}$
$⇒logy+\frac{x}{y}\frac{dy}{dx}=logx\frac{dy}{dx}+\frac{y}{x}$
$⇒(\frac{x}{y}-logx)\frac{dy}{dx}=\frac{y}{x}-logy$
$⇒(\frac{x-ylogx}{y})\frac{dy}{dx}=\frac{y-xlogy}{x}$
$∴\frac{dy}{dx}=\frac{y}{x}(\frac{y-xlogy}{x-ylogx})$