x^y+y^x=1

Answer 1

To differentiate u^v, where u and v are functions of x, use a combination of the easy formulas you already know:

y = u^n
y' = n u^(n-1) u'

y = a^v
y' = lna a^v v'

x^y + y^x = 1
x^(y-1) y' + lnx x^y + y^(x-1) y' + lny y^x = 0

(x^(y-1) + y^(x-1))y' = -lnx x^y - lny y^x

y' =

   lnx x^y + lny y^x
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nbsp;  x^(y-1) + y^(x-1)

Answer 2

Rats. How about

y x^(y-1) y' + lnx x^y + x y^(x-1) y' + lny y^x = 0

(y x^(y-1) + x y^(x-1))y' = -lnx x^y - lny y^x

Note that x^(y-1) = x^y/x, so

(y/x x^y + x/y y^x)y' = -lnx x^y - lny y^x

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