Answer:
[tex]\dfrac{dy}{dx} = \dfrac{xy\ln (y) -y}{xy\ln (x) - x}[/tex]
Step-by-step explanation:
Considering [tex]x^y = y^x[/tex] we want to find [tex]\dfrac{dy}{dx} = y'[/tex]
Once we have exponents I would just apply the natural logarithm both sides, therefore, by the basic property of logarithms, we know
[tex]x^y = y^x \implies y \ln(x) = x\ln (y)[/tex]
Now we can differentiate the equation and using the product rule, we have
[tex]$\dfrac{d}{dx}y \cdot\ln (x)+\dfrac{d}{dx}\ln (x) \cdot y = \boxed{y' \cdot\ln (x) + \dfrac{1}{x} \cdot y}$[/tex]
[tex]$\dfrac{d}{dx}x \cdot\ln (y)+\dfrac{d}{dx}\ln (y) \cdot x = \boxed{1 \cdot\ln (y) + \dfrac{x}{y} \cdot y'}$[/tex]
Thus,
[tex]y' \cdot\ln (x) + \dfrac{1}{x} \cdot y =\ln (y) + \dfrac{x}{y} \cdot y'[/tex]
[tex]\implies y' \cdot\ln (x) - \dfrac{x}{y} \cdot y' =\ln (y) - \dfrac{1}{x} \cdot y[/tex]
[tex]\implies y' \left(\ln (x) - \dfrac{x}{y}\right) =\ln (y) - \dfrac{y}{x}[/tex]
[tex]\implies y' = \dfrac{\ln (y) - \dfrac{y}{x}}{\ln (x) - \dfrac{x}{y} }=\dfrac{xy\ln (y) -y^2}{xy\ln (x) - x^2}[/tex]
