InterviewSolution
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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 )` |
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Answer» (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 )` |
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