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

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

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