यदि `y^(x) + x^(y) + x^(x) = a^(b) ` हो , तो `(dy)/(dx)` ज्ञात कीजिए ।

Explanation

` :. 1/u * (du)/dx = x d/dx log y + log y d/(dx) x `
` = x xx 1/y * dy/dx + log y xx 1`
` (du)/ dx = u [x/y * dy/dx + log y ] `
` = y^(x) [ x/y . (dy)/dx + log y ] `
` i/ v . (dv)/(dx) = y d/(dx) log x + log x d/ (dx) .y`
` = y xx 1/x + log x , (dy)/dx `
` :. (dv)/dx = v[ y/x + log x . dy/dx ]`
` = x^(y) [ y/x + log x . dy /dx]`
` log w = x log x `
` 1/ w , (dw) /(dx) = x d/(dx) log x + log x d/ (dx) x `
` = x xx 1/x +m log x xx 1 `
` :. (dw)/(dx) = w [ 1+ logx]`
`(du)/dx + (dv)/(dx) + (dW)/(dx) =0`
` y ^(x) [ x/y. (dy)/(dx) + log y ] + x^(y) [ y/x + log x. (dy)/(dx) ] + x^(x) [ 1 + log x] = 0 `
` x. y^(x-1) (dy)/(dx) + y^(3) log y + y . x^(y - 1) + x^(y) . log x (dy)/(dx) = - x^(x) [ 1+ log x]`
` (dy)/(dx) [ x.y^(x -1) + x^(y) log x ] = - [ y^(x) log y + y , x^(y-1) +x^(x) ( 1 + log x ) ] `
` (dy)/(dx) = -([y^(x) log y + y.x^(y-1) + x^(3) ( 1+ log x) ])/( x .y^(x-1) + x^(x) .log x )`