Question

Find $\frac{dy}{dx}$ where $x^y=y^x;x>0,y>0$.

Answer 1

Given, $x^y=y^x$

Taking $\log$ on both sides $y\times \ln x=x\times \ln y$

Now differentiating on both sides applying product rule,

$\Rightarrow \frac{dy}{dx}\ln x+\frac{y}{x}=\ln y+\frac{x}{y}\times \frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}(\ln x-\frac{x}{y})=\ln y-\frac{y}{x}$

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