Q:

Derivative of implicit function with exponential functions of each other

We have the equation:
$$
x^y = y^x +y
$$
Which defines an implicit function $y(x)$ at the point $(2,1)$. I'm asked to find the derivative at $y'(2)$.
I saw the answer in Wolfram:
$$
y'(x) = \frac{y (y x^y-x y^x \ln{y})}{x (-y x^y \ln{x}+x y^x+y)}
$$
Which gives $y'(2)=\frac{-1}{\ln{4}-3}$.
I don't understand how to get there. When I try to dervie, after taking $\ln$ from both sides I get:
$$
y'\ln{x} + \frac{y}{x} = \frac{y^x\ln{x} + 1}{y^x+y}y'
$$
$$
y'[\frac{1 - y\ln{x}}{y^x+y}] = \frac{y}{x}
$$
$$
y' = \frac{y^{x+1}+y^2}{x - xy\ln{x}}
$$

A:

By the property of derivatives, if we find the derivative of all individually, then we can add them all and do the implicit differentiation. 
$x^y=t$ 
$y\log{x}=\log t$ 
$t(\frac{dy}{dx}\log x+\frac yx)=\frac{dt}{dx}$ 
For $y^x=u$ 
$x\log y=\log u$ 
$u(\log y+\frac{x}{y}\frac{dy}{dx})=\frac{du}{dx}$ 
So $\frac{dt}{dx}=\frac{du}{dx}+\frac{dy}{dx}$ 
$x^y(\frac{dy}{dx}\log x+\frac yx)=y^x(\log y+\frac{x}{y}\frac{dy}{dx})+\frac{dy}{dx}$ 
Substituting the point 
$2(\frac{dy}{dx}\log2+\frac12)=\log1+2\frac{dy}{dx}+\frac{dy}{dx}$ 
$2\log2\frac{dy}{dx}+1=3\frac{dy}{dx}$ 
$\frac{dy}{dx}(3-\log4)=1$ 
$\frac{-1}{\log4-3}=\frac{dy}{dx}$. Which is the required result.

