यदि ` y= (1+(1)/(x) )^(x) + x^(1+(1)/(x))` तो `(dy)/(dx) ` का मान ज्ञात कीजिए|

Explanation

दिया है- ` " "y= (1+(1)/(x))^(x)+x ^(1+(1)/(x))`
माना ` " "u =(1+(1)/(x) ) ^(x)` तथा ` " "v=x^(1+(1)/(x))`
` rArr " "y=u +v " "rArr(dy)/(dx)= (du)/(dx)+(dv)/(dx)" "...(1)`
अब ` " "u= (1+(1)/(x))^(x) rArr log u =x log (1+(1)/(x))`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर
` " "(1)/(u) (du)/(dx) =x (d)/(dx) log (1+(1)/(x) )+log (1+(1)/(x))(d)/(dx) (x)`
` " "= x *(1)/(1+(1)/(x))*(d)/(dx) (1+(1)/(x)) +log(1+(1)/(x))*1`
`rArr " "(du)/(dx)=u [(x^(2)/x+1)(0-(1)/(x^(2)))+ log (1+(1)/(x))]`
` " "= (1+(1)/(x))^(x) [log (1+(1)/(2))-(1)/(1+x)] " "...(2)`
अब ` v=x^(1+(1)/(x))" "rArrlog v =(1+(1)/(x) ) log x `
दोनों पक्षों का x के सापेक्ष अवकलन करने पर
` rArr " "(1)/(v) (dv)/(dx) =(1+(1)/(x))*(1)/(x) +log x*(0-(1)/(x^(2)))`
` rArr " "(dv)/(dx)=v [(1)/(x) +(1)/(x^(2) )-(1)/(x^(2))log x ]`
` " "= x^(1+(1)/(2))[(1)/(x)+(1)/(x^(2))(1-log x )]`
समीकरण (1 ),(2 ),व (3 ) से,
` (dy)/(dx) = (1+(1)/(x))^(x) [log (1+(1)/(x))-(1)/(1+x)]`
` " "+ x^(1+(1)/(x))[ (1)/(x)+ (1)/(x^(2))(1-log x )]`