Swadesh Rath
Calculus
Level
4
if f(x) = 1/(1-x^2) , |x| < 1
so if nth derivative of this function f(x) is represented as ' tn '
For even n, find the value of tn at x = 0 ( in terms of n ).
2^n
(n)(n+1)/2
n!
n
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1 solution
if f(x) = 1/(1- x^2) , where |x| < 1 so we can write this function easily as - f(x) = 1 + x^2 + x^4 + x^6 .... infinite terms G.P. with common difference as x^2 and first term as 1 so when we differentiate it nth time ... all terms before x^n become zero and all after x^n which contain the term x - become zero ... so finally our only required term is X^n ... when we differentiate it n times ... we get n(n - 1)(n - 2) ... 3.2.1 = n!