InterviewSolution
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InterviewSolution
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`(log x)^(tan x ) ` का ` sin (mcos ^(-1) x ) `के सापेक्ष अवकल गुणांक ज्ञात कीजिए |
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Answer» माना ` y_1 =(log x ) ^(tan x ) ` तथा ` y_2 sin (mcos ^(-1) x )` ` because " "y_1 =(log x )_^(tan x ),` दोनों पक्षों का लघुगणक लेने पर, ` log y_1 =tan x log x ` ` therefore " "(1)/(y) (dy)/(dx) =(tan x )/(xlog x ) +sec ^(2) x log log x ` ` rArr " "(dy_1) /(dx) =(log x )^(tan x ) [(tan x )/(xlog x ) +sec ^(2) x log log x ]` तथा ` " "y_2 =sin (m*cos ^(-1) x ) ` ` therefore " "(dy_2)/(dx) =(d)/(dx) sin (mcos ^(-1) x )` या ` (dy_2)/(sx) =(-mcos (mcos ^(-1) x ))/(sqrt((1-x^(2))))` अतः` (dy_1)/(dy_2) =(y_1 ("का x के सापेक्ष अवकलन")) / (y_2("का x के सापेक्ष अवकलन"))` ` =((logx )^(tan x) [(tan x)/(xlog x ) +sec ^(2) xlog log x ])/(-(mcos (mcos ^(-1)x))/(sqrt(1-x^(2))))` ` =- ((log x)^(tan x) [(tan x)/(xlog x )+sec ^(2) x log log x ]sqrt(1-x^2))/(mcos (mcos ^(-1)x))` |
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