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 (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}`