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

Explanation

` " "y= (log x )^(x) +x^(log x ) `
माना ` " "y_1 =(log x )^(x) ` व् ` y_2 =x ^(log x )`
`therefore " "y= y_1 +y_2" "...(1)`
` " "(dy)/(dx) =(dy_1)/( dx)+ *( dy_2)/(dx) `
` " "y_1 =(log x )^(x) ` लेने पर,
` rArr " "log y_1 =x log (log x )`
x के सापेक्ष अवकलन करने पर
` rArr " "(1)/(y_1) *(dy_1) /(dx)x* (1)/(log x )* (1)/(x) +log (log x ) *1`
` " "=(1)/(log x) + log (log x ) `
` (dy_1)/(dx) = y_1[(1)/(log x ) +log (log x )]`
` " "= (log x )^(x) [(1)/(logx ) +log (log x ) ]" "....(2)`
अब ` y_(2) =x^(log x) `
` " " log y_2=(log x )(log x )= (log x)^(2)`
x के सापेक्ष अवकलन करने पर
` rArr " "(1)/(y_2 ) (dy_2)/(dx) =2log x *(1)/(2) `
` " " (dy_2) /(dx) =y_2 [2log x*(1)/(x) ]= x ^(log x) [(2log x )/(x)]`
` " "= x^(log x -1) * 2log x " "...(2)`
` " "` समीकरण )(1 ),(2 ) व् (3 ) से,
` (dy)/(dx) =(log x ) ^(x-1) +(log x ) ^(x) xx log (log x ) `