x^y=y^x --->
y log(x) = x log(y) ----->
log(x) dy + y/x dx =
log(y) dx + x/y dy ---->
[log(x) - x/y] dy = [log(y) - y/x] dx
---->
dy/dx = [log(y) - y/x]/[log(x) - x/y]
find y'(x) if x^y=y^x
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