1.

`x^(y) + y^(x) = 1`

Answer» `x^(y) + y^(x) = 1`
`rArr u+v = 1 " where "u=x^(y) " and "v=y^(x)`
`rArr (du)/(dx) + (dv)/(dx) = 0 " "....(1)`
`"Now " u = x^(y)`
`rArr "log"u = "log"(x^(y)) = y"log"x`
`rArr (1)/(u) (du)/(dx) = y * (d)/(dx)"log"x + "log"x * (d)/(dx)y`
`rArr (du)/(dx) = u[(y)/(x) + "log"x(dy)/(dx)]`
`=x^(y)[(y)/(x) + "log"x(dy)/(dx)] = y * x^(y-1) + x^(y) "log"x(dy)/(dx)`
`"and " v = y^(x)`
`rArr " log"v = "log"y^(x) = x"log"y`
`rArr (1)/(v) (dv)/(dx) = x (d)/(dx)"log" y + "log" y * (d)/(dx)x`
`rArr (dv)/(dx) = v[(x)/(y)(dy)/(dx) + "log"y * 1]`
`=y^(x)[(x)/(y)(dy)/(dx) + "log"y]`
`= x * y^(x-1) (dy)/(dx) + y^(x) "log"y`
From equation (1)
`y * x^(x-1) + x^(y)"log" x (dy)/(dx) + x * y^(x-1)(dy)/(dx) + y^(x) "log" y = 0`
`rArr (dy)/(dx) (x^(y)"log"x + x * y^(x-1))`
`=-(yx^(y-1) + y^(x)"log"y)`
`rArr (dy)/(dx) = -(y * x^(y-1) + y^(x) "log"y)/(x^(y) "log"x + x * y^(x-1))`


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