निम्न फलनों के x के सापेक्ष अवकलन कीजिये|
` (i) ((x+1)^(2)sqrt(x-1))/((x+4)^(e)e^x)`
` (ii) (xlog x ) ^(log log x )`

Explanation

(i ) माना ` y= ((x+1)^(2) (x+1)^(1//2))/((x+4)^(3)e^(x))`
दोनों पक्षों का लघुगणक लेने पर
` log y= log [((x+1)^(2) (x-1)^(1//2))/((x+4)^(3)e^(x))]`
` =log (x+1)^(2) +log (x-1)^(1//2) -log (x+4) ^(3) -log e^(x)`
या ` log y =2 log (x+1) +(1)/(2) log (x-1) -3 log (x+4) -x `
` " "(because log _e e=1)`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
` (d)/(dx) log y (dy)/(dx) =2 (d)/(dx) log (x+1) +(1)/(2) (d)/(dx) log (x-1)`
` " "-3(d)/(dx) log (x+4)-1`
` rArr (1)/(y) (dy)/((dx) )=(2)/(x+1) +(1)/(2(x-1))-(3)/(x+4)-1`
` rArr (dy)/(dx) =y [(2)/(x+1)+ (1)/(2(x-1))-(3)/(x+4) -1]`
` rArr (dy)/(dx)=((x+1)^(2) sqrt((x-1) ))/((x+4 )^(3) e^(x) )[ (2) /(x+1)+(1)/(2(x-1))-(3)/(x+4)-1`
`(ii)` माना ` y= (xlog x ) ^(log log x ) `
दोनों पक्षों का लघुगणक लेने पर,
`log y= log [(xlog x ) ^(log log x )]`
`" " =(log log x )log (xlog x )`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर
` (d)/(dy) log y (dy)/(dx) =[log log (x) ] (d)/(dx) log (xlog x ) `
` " "+ log (xlog x ) (d)/(dx) log log (x) `
` rArr " "(1)/(y) (dy)/(dx) =log log(x) [(1) /(xlog x ) {x(d) /(dx) log x +log x (d)/(dx) x }]`
` " "= + log (x log x ) [ (1)/(x log x )]`
` rArr (1)/(y)(dy)/(dx) =log log (x) [(1) /(xlog x ) (1+ log x ) ]`
` " "+ log (xlog x ) [(1) /(xlog x )]`
` rArr " "(dy)/(dx) =(xlog x )^(log log x ) [(log log (x))/(x log x )](1+log x ) `
` " " + log (x log x ) /( xlog x )`