Calculus :)
posted by Hailey .
find y'(x) if x^y=y^x

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] 
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derivative x^y=y^x
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