Did I take the derivative correctly? $x^y=y^x$ Need to differentiate following equation:
$$x^y=y^x$$
My attempt:
$$x^y\log(x) \cdot y' = y^x\log(y)\cdot1;  $$
$$y'=\frac{y^x\log(y)}{x^y\log(x)}$$
Please tell me if I've made a mistake.
 A: Just to let you know where you went wrong: it was in differentiating the right hand side. You're trying to find 
$
\frac{d}{dx} y^x
$
and applying the rule that 
$
\frac{d}{dx} a^x = a^x \ln (a),
$
with $y$ replacing $a$. Unfortunately, that "rule" applies only in the case where $a$ is a constant; in your problem, the base $y$ is actually a function of $x$, and so the rule doesn't apply. Thus to properly compute the derivative, you have to rewrite $y^x = e^{x \ln (y)}$, as others have already shown. 
A: Not quite. The idea is to take the $\log$ of both sides and implicitly differentiate. $$ \begin {eqnarray*} x^y &=& y^x \\ y \cdot \log (x) &=& x \cdot \log (y) \\ y' \cdot \log(x) + \dfrac {y}{x} &=& \dfrac {x}{y} \cdot y' + \log (y) \\ y' \cdot \left( \log(x) - \dfrac {x}{y} \right) &=& \log(y) - \dfrac {y}{x} \\ y' &=& \dfrac {\log (y) - \dfrac {y}{x}}{\log (x) - \dfrac {x}{y}}, \end {eqnarray*} $$so our answer is $$ y' = \boxed {\dfrac {x \cdot \log (y) - y}{y \cdot \log (x) - x} \cdot \dfrac {y}{x}}. $$
A: $ y^x = x^y $
$ \ln y^x = \ln x^y$
$x \ln y = y \ln x $
$\ln y + (x/y)(dy/dx) = (dy/dx)(ln x) + y/x $
$(x/y)(dy/dx) - (dy/dx)(\ln x) = y/x - \ln y $
$(dy/dx)(x/y - \ln x) = y/x - \ln y $
$(dy/dx) = (y/x - \ln y)/(x/y - \ln x)$
$(dy/dx) = (y^2 - xy\ln y)/(x^2 - xy\ln x)$
