Question

find $\frac{dy}{dx}$

$x^{y} . y^{x} =1$

Answer 1

$x^y\ y^x = 1\\\\Log\ x^y\ y^x=0\\\\y\ Log\ x+ x\ log\ y=0\\\\dy/dx\ Log\ x+1/x\ y+x/y\ dy/dx\ +Log\ y=0\\\\\frac{dy}{dx}[Log\ x+\frac{x}{y}]=-Log\ y-\frac{y}{x}\\\\\frac{dy}{dx}[1-Log\ y]\frac{x}{y}=\frac{y}{x}[Log\ x - 1]\\\\\frac{dy}{dx}=\frac{y^2\ (Log\ x/e)}{x^2\ (Log\ e/y)}$