Derivative of $x^y=y^x$ defines: $y=y(x)$ I need to find the derivative. given that:
$$x^y=y^x$$
defines:
$$y=y(x)$$
Thank you!
 A: Consider $y=y(x)$ and use implicit differentiation on:
$$x^{y(x)}=y(x)^x$$
Carefully computing the derivative (don't forget the chain rule) gives:
$$x^y \left( \color{blue}{y'} \ln x + \frac{y}{x} \right) = y^x \left( \ln y + \frac{x\color{blue}{y'}}{y} \right)$$where I omitted the argument $x$ of $y$ and $y'$ to simplify the notation. Now solve for $\color{blue}{y'}$, this only requires some algebraic manipulations. You can check your answer here.
A: Method 1
\begin{align*}
  y^{x} &= x^{y} \\
  x\ln y &= y\ln x \\
  \ln y+\frac{xy'}{y} &= y'\ln x+\frac{y}{x} \\
  \left( \frac{x}{y}-\ln x \right)y' &= \frac{y}{x}-\ln y \\
  y' &= \frac{y(y-x\ln y)}{x(x-y\ln x)}
\end{align*}
Method 2
Let $y=(t+1)x$, then
\begin{align*}
  [(t+1)x]^{x} &= x^{(t+1)x} \\
  (t+1)^{x}x^{x} &= x^{tx} x^{x} \\
  (t+1)^{x} &= x^{tx} \\
  t+1 &= x^{t} \\
  x &= (t+1)^{1/t} \\
  y &= (t+1)^{(t+1)/t} \\
  \dot{x} &= (t+1)^{1/t-1} \times \frac{t-(t+1)\ln (t+1)}{t^{2}} \\
  \dot{y} &= (t+1)^{1/t+1} \times \frac{t-\ln (t+1)}{t^{2}} \\
  \frac{dy}{dx} &= (t+1)^{2} \frac{t-\ln (t+1)}{t-(t+1)\ln (t+1)} \\
\end{align*}
Further points:
We omit the trivial case $y(x)\equiv x$ here.
