Asked by Neeraj kumar
                x^y+y^x=1
            
            
        Answers
                    Answered by
            Steve
            
    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
- ----------------------
nbsp; x^(y-1) + y^(x-1)
    
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
- ----------------------
nbsp; x^(y-1) + y^(x-1)
                    Answered by
            Steve
            
    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
...
    
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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