Question

Find Dy/dX when x^y×y^x=1​

Answer 1

by dy/dx, we'll be assuming that "y" is encapsulating a function, a function in terms of "x".

$x^y\cdot y^x=1\implies \stackrel{\textit{\large product rule}}{\stackrel{\textit{chain rule}}{yx^{y-1}(1)}\cdot y^x+x^y\cdot \stackrel{\textit{chain rule}}{xy^{x-1}\cdot \frac{dy}{dx}}}~~ = ~~0 \\\\\\ y^{1+x}x^{y-1}+x^{y+1}y^{x-1}\cdot \cfrac{dy}{dx}=0\implies \cfrac{dy}{dx}=\cfrac{-y^{1+x}x^{y-1}}{x^{y+1}y^{x-1}} \\\\\\ \cfrac{dy}{dx}=-~~ y^{(1+x)-(x-1)}~~x^{(y-1)-(y+1)} \\\\\\ \cfrac{dy}{dx}=-~~ y^2x^{-2}\implies \cfrac{dy}{dx}=\cfrac{-y^2}{x^2}$