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Differential Calculus

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 Joined: 14 May 2007 Post: 6
15 May 2007 15:43:30 IST
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Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Differential Calculus

show that the nth derivative of  (1-x2 ) is

n! /2( 1/(1-x)n+1) + (-1)n /(1+x)n+1 )

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Joined: 27 Dec 2006
Posts: 565
15 May 2007 22:11:40 IST
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the question is wrong it should be
nth derivative of 1/(1-x2)
let y= 1/(1-x2)
resolving the above expression in partial fractions we get
y=1/2[1/(1+x)+1/(1-x)]

now differentiating n times we get
y=n! /2( 1/(1-x)n+1) + (-1)n /(1+x)n+1 )

Blazing goIITian

Joined: 12 Nov 2006
Posts: 413
15 May 2007 22:32:02 IST
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hy shakirshafi......can u plzzzzz tell me da general formula for claculating da nth derivative

Blazing goIITian

Joined: 12 May 2007
Posts: 328
15 May 2007 23:06:53 IST
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general formula of nth derivative of 1/(ax+b)^r   is
(-1)^n  (n+r-1)! (a)^n / (ax+b)^(n+r) (r-1)!
space indicates multiplication sign
it is a very important formula and very few people know abt it
try solving with this formula n u will make wonders

Hey I really deserve rating yaar

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Joined: 27 Dec 2006
Posts: 565
16 May 2007 15:19:01 IST
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well  general method i can give u a very good idea
here

suppose a function be xm
then the nth derevative will be like this see

y=xm
y`=mxm-1
y``=m*(m-1)xm-2
for the nth derevative
dny/dxn=m*(m-1)(m-2)(m-3)(m-4)............(m-n)xm-n
we multiply and divide by n!
dny/dxn=m!/n!xm-n
dny/dxn=mPn xm-n
this is the method
using this we can calculate for

general formula of nth derivative of (ax+b)-r   is
(-1) (n+r-1)! (a)n / (ax+b)n+r (r-1)!

for such problems we need to find a relation between the derevatives!!
as a function of the number of times differenciated!!

these things dont come in JEE!!

all this successive differenciation and leibnitz theorum
i.e sucessive differenciation of product of two functions!!

New kid on the Block

Joined: 14 May 2007
Posts: 6
19 May 2007 10:03:08 IST
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sorry for givin a wrong question
shakirshafil gave the rite question and answer

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