InterviewSolution
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InterviewSolution
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निम्न फलनों का अवकलन कीजिये| `(i) x^(x)` ` (ii) x^(sin-1)x` `(iii) (sin x )^(tanx )` |
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Answer» (i ) माना `" "y =x^(x)` दोनों पक्षों का लघुगणक लेने पर ` " "log y= log (x^(x))=x log x ` दोनों पक्षों का x के सापेक्ष अवकलन करने पर ` " "(d)/(dy) logy(dy)/(dx) =(d)/(dx) [x log x]` ` rArr " "(1)/(y)(dy)/(dx) =x(d)/(dx) log x +log x (d)/(dx) x ` ` rArr " "(1)/(y)(dy)/(dx) =x*(1)/(x) +log x` `rArr" "(dy)/(dx) =y (1+log x)` अतः ` " "(dy)/(dx) =x^(3) (1+log x ) " "(because y= x^(x))` `(ii)` माना ` " "y=x ^(sin -1)x ` दोनों पक्षों का लघुगुणक लेने पर, ` " "log y =log x ^(sin-1x)` ` " "log y =sin ^(-1) xlog x ` ` therefore (d)/(dy) log y (dy)/(dx)=sin ^(-1) x(d)/(dx) log x +log x(d)/(dx) sin ^(-1)x` ` rArr " "(1)/(y)(dy)/(dx) =(sin ^(-1)x)/(x) +(log x)/(sqrt(1-x^(2)))` ` rArr " "(dy)/(dx)= y[(sin ^(-1) x)/(x)+ (log x)/(sqrt(1-x^(2))]]` ` therefore" "(dy)/(dx) =x^(sin -1_x)[(sin ^(-1)x)/(x)+(log x)/(sqrt(1-x^(2)))]` ` " " (because y= x^(sin ^(-1_x)))` `(iii)` माना ` " " y= (sinx )^(tan x)` दोनों पक्षों का लघुगुणक लेने पर `" "log y=log (sinx )^(tan x ) ` ` " "=tan x log (sinx ) ` दोनों पक्षों का x के सापेक्ष अवकलन करने पर ` (d)/(dy) log y (dy)/(dx) =tan x (d)/(dx) log sin x+ log sin x (d)/(dx) tan x ` ` rArr " "(1)/(y) (dy)/(dx) =tan x cot x +sec ^(2) x log sin x ` ` " "= 1+sec ^(2) x log sin x ` ` rArr " "(dy)/(dx) =y [1+sec^(2) x log sin x ]` ` " "= (sin x ) ^(tan x ) [1+sec^(2)x log sin x ]` |
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