यदि ` x^(x)+x^(y)+y^(x) = a ^(b) ,` तब `(dy)/(dx)` का मान ज्ञात कीजिए|

Explanation

`" " x^(x) + x^(y) +y^(x) =a^(b) `
माना ` y_1 =x^(x), y_2 =x^(y) ` व् ` y_3 =y^(x)` तब
` y_1 +y_2 +y+3 =a^b`
अब `" "y_1 =x^(x)`
` " "log y_1 =x log x `
x के सापेक्ष अवकलन करने पर
` (1)/(y_1) *(dy_1)/(dx) =x*(d)/(dx) (log x )+ (log x ) +(log x )(d)/(dx) (x)`
` " "x*(1)/(x) + log x *1 `
` (dy_1)/(dx) =y_1 p[1+ log x ] =x^(x) [1+log x ]`
अब,` " "y_2 =x^(y)`
` log y_2 =ylog x `
x के सापेक्ष अवकलन करने पर
` (1)/(y_2) *(dy_2)/(dx)=y (d)/(dx) (log x )+(logx )(d)/(dx) (y)`
` " "=[y*(1)/(x) +(log x )(dy)/(dx) ]`
` (dy_2) /(dx) =y_2 [(y)/(2) +(lo gx) (dy)/(dx) ] =x^(y) [(y)/(x) +(log x ) (dy)/(dx) ]`
और ` y_3 =y^(x) `
` log y_3 =x log y`
x के सापेक्ष अवकलन करने पर
` (1)/(y_3) (dy_3)/(dx) =x *(d)/(dx) (log y)+(log y )*(d)/(dx) (x) `
`" "=[ x*(1)/(y) *(dy)/(dx ) +(log y) *1]`
`(dy_3)/(dx) =y_3 [(x)/(y) *(dy)/(dx) +log y] =y^(x) [(x)/(y) *(dy)/(dx) +log y]`
समीकरण (1 ) से
`y_1 +y_2 +y_3 =a^b `
` rArr *(dy_1) /(dx) +(dy_2)/(dx) +(dy_3)/(dx) =0`

`x^(x) (1+log x )+ x^(y) {(y)/(x)+ (log x ) (dy)/(dx) }+ y^(3){(x)/(y) (dy)/(dx) +(logy)}=0`
` rArr {x^(x) (1+log x )+ y*x^(y-1)+y^(x) (log y)}+ {x^(y)(log x)+xy^(x-1) }*(dy)/(dx) =0`
` rArr (dy)/(dx) =(-{x^(x)(1+log x )+ y*x ^(y-1) + y^(x) (log y)})/({x^(y)(log x )+xy^(x-1)})`