Question

Find $\frac{dy}{dx}$, if $x^y+y^x=1$

Answer 1

Let $b(x.y)=x^y+y^x-1$

$b^{\prime }=0$

$b^{\prime }xdx+b^{\prime }ydy=0$

$b^{\prime }x+b^{\prime }y\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-b^{\prime }x}{b^{\prime }y}=-\frac{y{x}^{y-1}+y^{x}10y}{x{y}^{x-1}+x^{y}10x}$.