Jtdylktuy

I know how to derivate and I've found the implicit differentiation of $y^{sin x} = x^{sin y}$ which is $y' = frac{frac{sin x}{y} - cos x ln y}{frac{sin y}{x} - cos y ln x}$, but how do I obtain $y$ alone, is there a way?

PD: Thanks you all for helping me understand this problem :)

Apparently $sin x=0$ or $sin y=0$ doesn't satisfy the equation.

And $x=y>0$ are solutions to the equation. We now only consider solutions in which $xne y$.

Assuming $x,y>0$, take logarithm on both side, we have

$sin x ln y = sin y ln x$

or

$cfrac {ln x}{sin x}=cfrac{ln y}{sin y}$

Define $f(x)=cfrac {ln x}{sin x}$

Plot the graph of f(x), test it with $y=c$ ($c in mathbb{R}$), provided the graph intersect $y=c$ for more than once, the corresponding $x$ values of the intersections form a set ${x_i}, ige 2$, $x=x_i , y=x_j, ine j$ is a solution to the original equation with $xne y$.

PS: for example, the line $y=10$ will intersect the graph of our $f(x)$ over many dozens of times, let's say it's $k$ times; name the x-coordinates of these intersection $x_1, x_2, cdots, x_k$. Apparently $f(x_1)=f(x_2)=cdots=f(x_k)$, which implies $cfrac{ln x_1}{sin x_1}=cfrac{ln x_2}{sin x_2}=cdots=cfrac{ln x_k}{sin x_k}$ which in turn means $x=x_1, y=x_2$, etc (Pick $2$ out of $k$ and assign them to $x,y$) are all solutions to the original equation. You will have to find the exact values numerically though.

If you plot the $(x,y)$ pairs for which this holds, you'll see that this won't only be problematic for $x=0$
$[0,8pi]$]">

By symmetry, the equation $y^{sin x} = x^{sin y}$ is satisfied if
$$y=x.$$

EDIT:

There might be more solutions, for instance:

If $x=0$, the equation for $y$ is $y^0=1=0^{sin y}$ which holds for $sin y = 0$ and thus $y = pi n$ with $ninmathbb{Z}$.

It is the same as
$$x^{1/sin(x)}=y^{1/sin(y)}$$
You can plot $x^{1/sin x}$ on Wolfram Alpha.

That function seems to be increasing between $0$ and $pi$, so there are no solutions below $pi$, but there seem to be solutions between $0$ and $2pi$.

We'll only consider the equation when $x$ and $y$ are both positive. (The expression $x^{sin y}$ is, in general, not well defined when $x$ is negative, and likewise for $y^{sin x}$.)

For $x,ygt0$, the equation is clearly satisfied if $sin x=sin y=0$, since both sides become equal to $1$. Thus the solution set includes the set ${(mpi,npi)mid m,ninmathbb{N}}$.

If $sin xsin ynot=0$, then $x^{sin y}=y^{sin x}$ is equivalent to $x^{1/sin x}=y^{1/sin y}$. Setting aside the interval $(0,pi)$, it's easy to see that, on any interval of the form $(2mpi,(2m+1)pi)$, the function $f(x)=x^{1/sin x}$ varies continuously from $infty$ down to $(2m+1/2)pi$ and then back up to $infty$, while on intervals of the form $((2m-1)pi,2mpi)$, it varies continuously from $0$ up to $((2m-1/2)pi)^{-1}$ and then back down to $0$. (Note, the function doesn't actually take the values $infty$ or $0$; those are just the appropriate limits at the endpoints.) Consequently, almost every value in the range of $f$ is achieved multiple times (and the closer the value is to $infty$ or $0$, the more often it is achieved). This implies a continuum of solutions to $x^{sin y}=y^{sin x}$, in agreement with the graph in Ákos Somogyi's answer.