Understanding the graph for $x^y = y^x$ I graphed $x^y=y^x$ and it is a union of the line y=x, with some other curve.  So my first question is, how do I derive that other curve? My next question is, why don't I get the same graph when I plot $x^{1/x}=y^{1/y}$?
 A: Let $y=\mathcal{R}x$, $\mathcal{R}$ a variable. Then,
$$x^{\mathcal{R}x}=(\mathcal{R}x)^x\\
\implies \mathcal{R}x\ln x=x\ln (\mathcal{R}x)=x\ln \mathcal{R}+x\ln x\\
\implies (\mathcal{R}-1)\ln x=\ln \mathcal{R}$$
We can choose $\mathcal{R}=1$, or, if $\mathcal{R}>1$,
$$x=\exp\left(\dfrac{\ln \mathcal{R}}{r-1}\right)=\mathcal{R}^{1/(\mathcal{R}-1)}$$
I.e.,
$$y=\mathcal{R}^{\mathcal{R}/(\mathcal{R}-1)}$$
The non-straight graph is the graph for $y$ for which $\mathcal{R}\neq 1$.
Since $\mathcal{R}x\neq \dfrac{1}{x}$, you don't get the same graph.
A: Here is the way I first saw it,
many years ago:
If $x^y = y^x$
with $x > 0$ and $y > 0$,
let $r = y/x$,
so $y = rx$.
We will derive a parameterization
for $x$ and $y$
in terms of $r$.
Then
$x^{rx} = (rx)^x$
or,
taking the $x^{th}$ root,
$x^r = rx$
or
$x^{r-1} = r$
or,
if $r \ne 1$,
$x = r^{1/(r-1)}$
and
$y = rx = r^{1+1/(r-1)}
= r^{r/(r-1)}$.
Letting $r$ go through
the reals $> 1$
gives point $(x, y)$ with
$x^y = y^x$.
Another way to look at this
is to consider the curve
$v = u^{1/u}$.
For each $v$ such that
$1 < v < e$,
there are two values of $u$,
$u_1 < e < u_2$,
such that
$u_1^{1/u_1} = v = u_2^{1/u_2}$.
Note that
if $1/(r-1)$ is an integer,
say $n$,
then
$r = 1+1/n$
and
$x = (1+1/n)^n$
and
$y = (1+1/n)^{n+1}$
are two rational numbers
such that
$x^y = y^x$.
$n=1$ gives $x=2$ and $y=4$ (the well known solution).
$n=2$ goves
$x = (3/2)^2 = 9/4$
and $y = (3/2)^3 = 27/8$
(it is definitely less well known that
$(9/4)^{27/8} = (27/8)^{9/4}$.
I believe that
these are the only rational solutions to
$x^y = y^x$,
but I do not have a proof.
What the heck, I'll make it a problem and,
I hope, harvest some points.
A: An alternative parameterization, which is a little more symmetric and found in the SE link provided in the comments by @SteveKass, can be derived as follows: $$x^y=y^x \\ x=a^b\quad y=a^c \\ a^cb\ln a=a^bc\ln a\implies a^cb=a^bc \\ a^{c-b}=\frac{c}{b} $$ Now we can let $c=b+1$ so that the equation becomes $$a=1+\frac{1}{b}$$ Therefore a parameterization for the second curve is $$\mathbf{x}(t)=\left(\left(1+\frac{1}{t}\right)^t\,,\,\left(1+\frac{1}{t}\right)^{t+1}\right)$$
As to why you're getting different graphs, it almost certainly is due to numerical errors in the CAS you're using. I don't know much about that area though, so I couldn't tell you what was causing the various problems.
A: Here is an analysis of what the graph looks like.  I think it shows that the Maple graph (see comments on the OP) is correct.  I haven't seen the Wolfram graph so can't comment on that.
Lemma.  Let $c$ be a real number and consider the equation
$$\frac{\ln x}{x}=c$$
for $x>0$.  The equation has


*

*one solution if $c\le0$, and this solution satisfies $x\le1$;

*two solutions if $0<c<e^{-1}$, and the solutions satisfy $1<x<e$ and $x>e$ respectively;

*one solution $x=e$ if $c=e^{-1}$;

*no solution if $c>e^{-1}$.


Proof.  Look at the graph of $(\ln x)/x$.
Corollary.  The equation $x^y=y^x$ has no solutions with $x\le1$ or $y\le1$, except for $x=y$.
Proof.  We have
$$\frac{\ln x}{x}=\frac{\ln y}{y}\ ;$$
if $x\le1$ then $LHS\le0$; by the lemma there is a unique value for $y$, and clearly it is $y=x$.
Theorem.  For any $x>1$ there is a unique $y\ne x$ such that $x^y=y^x$.  Moreover, if $1<x_1<x_2$ then the corresponding values $y_1,y_2$ satisfy $y_1>y_2$.  That is, the curve
$$x^y=y^x\ ,\quad x>1\ ,\quad y>1\ ,\quad x\ne y$$
is decreasing.  Finally, if $x\to1^+$ then $y\to\infty$, and if $x\to\infty$ then $y\to 1$; so the curve has asymptotes at $x=1$ and at $y=1$.
Proof.  If $x>1$ then once again we have
$$\frac{\ln x}{x}=\frac{\ln y}{y}\ ;$$
in this case $0<LHS<e^{-1}$ and so the lemma shows that there are two solutions for $y$.  But one of them is $y=x$, so only the other one satisfies $y\ne x$.  And if we look again at the graph of $(\ln x)/x$ we see that if $1<x_1<x_2\le e$ then $e\le y_2<y_1$; if $1<x_1<e<x_2$ then $y_2<e<y_1$; if $e\le x_1<x_2$ then $y_2<y_1\le e$; if $x\to1$ then $y\to\infty$; and if $x\to\infty$ then $y\to1$.  This completes the proof.
A: The "top" part of the graph (the left side of the curve, and the right side of the straight line) is the limit of:
$$
y=x \log_x\left(x \log_x\left(\log_x\left(x \log_x\left(\cdots\right) \right)\right) \right)
$$
The "bottom" part of the graph (the left side of the straight line, and the right side of the curve) is:
$$
y=\sqrt[x]{x^{\sqrt[x]{x^{\sqrt[x]{x^{\sqrt[x]{x^{\cdots}}}}}}}}
$$
This could be of some use in numerically calculating it I suppose.  But they converge very very slowly around x=e.  Here's an approximation:
Approximation of y^x=x^y as a pair of functions
