Question

Find $\frac{dy}{dx}$of the functions given in question.

$y^x=x^y.$

Answer 1

Given, $y^x=x^y.$

Taking log on both sides, we get $log{y}^x=log{x}^y\Rightarrow xlogy=ylogx$

differentiating both sides w.r.t. x, we obtain

$\frac{d}{dx}\left(xlogy\right)=\frac{d}{dx}\left(ylogx\right)\Rightarrow x\left(\frac{1}{y}\right)\frac{dy}{dx}+\left(logy\right)y\frac{1}{x}+\left(logx\right)\frac{dy}{dx}\left(Usingproductrule\right)\Rightarrow \frac{x}{y}\frac{dy}{dx}-\left(logx\right)\frac{dy}{dx}=\frac{y}{x}-logy\Rightarrow (\frac{x}{y}-logx)\frac{dy}{dx}=\frac{y}{x}-logy\Rightarrow (\frac{x-ylogx}{y}\frac{dy}{dx}=\frac{y-xlogy}{x})\Rightarrow \frac{dy}{dx}-\frac{y}{x}\left(\frac{y-xlogy}{x-ylogx}\right)$