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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 Feb 2007 21:22:23 IST
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differentiate the following :
1
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(x-3) (x-2) (x-1)
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i ain't a quitter! |
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12 Feb 2007 21:28:35 IST
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I think its 12x - 3x^2 - 11 divided by (x-1)^2 (x-2)^2 (x-3)^2 Pls confirm !!!
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Umang |
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12 Feb 2007 21:37:41 IST
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i believe the ans is -1/[2(x-3)^2] - 1/[2(x-2)^2] - 1/[(x-1)^2] plzzzzz tell me if i'm wrong newayzz gimme ur votes
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
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12 Feb 2007 22:17:14 IST
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Hey toshi , can u pls tell the correct answer ???
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Umang |
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12 Feb 2007 22:23:25 IST
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hey umang i tried solvin my equation and i got a biquadratic expression....... can u plzzzzz work it out
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
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12 Feb 2007 22:30:06 IST
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Hey vasanth , I solved the denominator and then applied the quotient rule of differentiation . I hope my method is correct ? Pls check and reply !!!!
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Umang |
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12 Feb 2007 22:34:02 IST
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hey umang i'd b greatful if u solve it here 4 me.... i'm not able to get ur solution..... plzzzz write it down so tat i can correct my mistake if i'm wrong..... cheeerssss
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12 Feb 2007 23:17:58 IST
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take logarithm and then solve '!!! it's damn easy then !
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13 Feb 2007 08:41:56 IST
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f(x) = f1(x) * f2(x) * f3(x)
f'(x) = f1'(x) * f2(x) * f3(x) + f1(x) * f2'(x) * f3(x) + f1(x) * f2(x) * f3'(x)
hence the answer
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i promise a salute for everyone who correctly answers my question. |
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13 Feb 2007 12:07:03 IST
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Let y= 1/[(x-1)(x-2)(x-3)] logy= -[log(x-1)+log(x-2)+log(x-3)] (1/y)dy/dx= -[1/(x-1) +1/(x-2)+1/(x-3)] dy/dx= -[1/(x-1) +1/(x-2)+1/(x-3)] y dy/dx= -[(x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)] dy/dx=-[3x2 -12x +11] Ans
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13 Feb 2007 12:16:00 IST
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well, vasant is almost right, jst li'l mistakes of sign!!! converting the given fraction into partial functions.... y = 1/[(x-3) (x-2) (x-1)] = [1/ { 2 (x-3) }] - [1/ {x-2}] + [1/ { 2 (x-1) }] Now we can differntiate it easily, y' = [-1/ { 2 (x-3)2 }] + [1/ {x-2}2 ] - [1/ { 2 (x-1)2 }]
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13 Feb 2007 15:11:01 IST
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If y= 1/[(x-1)(x-2)(x-3)] Taking log on both the sides we obtain lny= -[ln(x-1)+ln(x-2)+ln(x-3)] differentiating both the sides w.r.t x we obtain (1/y)dy/dx= -[1/(x-1) +1/(x-2)+1/(x-3)] dy/dx= -[1/(x-1) +1/(x-2)+1/(x-3)] y dy/dx = -[(x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)] dy/dx = -[3x2 -12x +11]
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13 Feb 2007 15:18:50 IST
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Both firdous and teja have outlined the correct approach, but the one used by edision (same as firdous) is the best.
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Sudeep Kumar
(B tech, IITd)
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13 Feb 2007 20:38:39 IST
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hey all ! Please tell the mistake in my solution . I have solved the denominator and then applied quotient rule . But the answer I am getting is different . I am getting (x-1)^2 (x-2)^2 (x-3)^2 in the denominator . Numerator is same . Pls explain !!!!!!!
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Umang |
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15 Feb 2007 20:45:24 IST
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