`(dy)/(dx)` ज्ञात कीजिए ।
`x^(y)+y^(x)=1`

Explanation

`x^(y)+y^(x)=1`
`rArr" "u+v=1" जहाँ " u=x^(y) " तथा "v=y^(x)`
`rArr" "(du)/(dx)+(dv)/(dx)=0" …(1)"`
अब `u=x^(y)`
`rArr" "logu=log(x^(y))=y log x`
`rArr" "(1)/(u)(du)/(dx)=y.(d)/(dx)logx+logx.(d)/(dx)y`
`rArr (du)/(dx)=u[(y)/(x)+logx(dy)/(dx)]`
`=x^(y)[(y)/(x)+logx(dy)/(dx)]=y.x^(y-1)+x^(y)logx(dy)/(dx)`
तथा `v=y^(x)`
`rArr" "log v=logy^(x)=x log y`
`rArr" "(1)/(v) (dv)/(dx)=x(d)/(dx)logy+logy.(d)/(dx)x`
`rArr" "(dv)/(dx)=v[(x)/(y)(dy)/(dx)+logy.1]`
`=y^(x)[(x)/(y)(dy)/(dx)+logy]`
`=x.y^(x-1)(dy)/(dx)+y^(x)logy`
समीकरण (1 ) से
`y.x^(y-1)+x^(y)log x(dy)/(dx)+x.y^(x-1)(dy)/(dx)+y^(x)logy=0`
`rArr" "(dy)/(dx)(x^(y)logx+x.y^(x-1))`
`=-(yx^(y-1)+y^(x)logy)`
`rArr" "(dy)/(dx)=-(y.x^(y-1)+y^(x)logy)/(x^(y)logx+x.y^(x-1))`