InterviewSolution
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InterviewSolution
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`sqrt((3x-4))` का x के सापेक्ष प्रथम सिद्धांत से अवकल गुणांक ज्ञात कीजिएः |
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Answer» `y=sqrt(3x-4)=(3x-4)^(1//2)` .......(i) `Rightarrow y+deltay=[3(x+deltay)-4]^(1//2)` `y+deltay=[(3x-4)+3deltax]^(1//2)........(ii)` `y+deltay-y=[(3x-4)+3deltax]^(1//2)-(3x-4)^(1//2)` ` =(3x-4)^(1//2) [(1+(3deltax)/(3x-4))^(1//2)]-(3x-4)^(1//2)` ` Rightarrow deltay=(3x-4)^(1//2) [(1+(3deltax)/(3x-4))^(1//2)-1]` ` Rightarrow deltay=(3x-4)^(1//2) [1+(1)/(2)((3deltax)/(3x-4))+((1)/(2)((1)/(2)-1))/(2!)((3deltax)/(3x-4))^2+.....oo-1]` ` =(3x-4)^(1//2) [(1)/(2)((3deltax)/(3x-4))+((1)/(2)((1)/(2)-1))/(2!)((3deltax)/(3x-4))^2+.....oo]` ` =(3x-4)^(1//2) .(1)/(2)(3deltax)/(3x-4)[1+((1)/(2)((1)/(2)-1))/(2!)((3deltax)/(3x-4))+.....oo]` ` =(3deltax)/(2(3x-4)^(1//4))[1+((1)/(2)((1)/(2)-1))/(2!)((3deltax)/(3x-4))+.....oo]` दोनों पक्षों `deltax` को से भाग देकर `underset(deltax to 0)` लेने पर, `underset(deltax to 0)lim (deltay)/(deltax)=(dy)/(dx)=underset(deltax to 0)lim (3)/(2(3x-4)^(1//2))=(deltax)/(deltax)[1+(((1)/(2)-1))/(2!)(3deltax)/((3x-4))+......oo]` `=3/(2(3x-4)^(1//2)) [1+0+0+.....]` `=(3)/(2sqrt((3x-4)))` `therefore (d)/(dx)(3x-4)^(1//2)=(3)/(2sqrt((3x-4)))` |
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