InterviewSolution
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निम्न फलन का x के सापेक्ष अवकलन कीजिये|` (i) x^(x) +(sin x )^(sin x)` |
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Answer» माना ` " "y= x^(x) +(sin x )^(sin x )` माना ` " " y_1 =x^(x) ` तथा ` y_2 =(sin x)^(sin x ) ` ` therefore " " y=y_1 +y_2 " "...(1)` punha ` " "y_1 =x^(x) ` ` therefore " "(dy_1)/(dx) =x^(x) [1+log x ] ` [udahran (1 )(i ) से ] तथा ` " "y_2 =(sin x)^(sin x ) ` दोनों पक्षों का लघुगणक लेने पर, ` log y_2 =sin x log sin x ` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, ` (d)/(dy_2) log y_2""(dy_2)/(dx) =sin x(d)/(dx) log sin x+ log sin x (d)/(dx) sin x ` ` rArr " "(1)/(y_2) (dy_2) /(dx) =cos x+ cos x log sin x ` ` " " =cos x (1+log sin x ) ` ` rArr " "(dy_2)/(dx) =y_2 cosx (1+log sin ) ` ` " "= cos x (sinx ) ^(sin x) [ 1+ log sin x ] " "...(3)` समी० (1 ) ,(2 ) व् (3 ) से, ` (dy)/(dx) =x^(3) (1+log x ) + cos x (sin x ) ^(sinx) [1+ log sin x ]` |
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