$$x^y = y^x$$
find an expression for $dy/dx$ in terms of $y$ and $x$
for the first part is $yx^{y-1}$ and for the second part i got $\ln(y)y^xdy/dx$
when solving for dy/dx i got $yx^{y-1}) / (\ln(y)y^x)$
Just was wondering if my work is correct?
Sorry for the bad formatting, Im not sure how to make it formatted nice.
It might be easier to take $\log$ first and work with the expression. We have $$y \log x = x \log y$$ Now differentiating we get $$y' \log x +\dfrac{y}x = \log y + x \dfrac{y'}y \implies y' \left(\log x - \dfrac{x}y \right) = \left(\log y - \dfrac{y}x \right)$$ Hence, we get that $$y' = \dfrac{\left(\log y - \dfrac{y}x \right)}{\left(\log x - \dfrac{x}y \right)}$$