Question

The nth derivative of the function $f\left(x\right)=\frac{1}{1-x^2}\left[wherex\in \left(-1,1\right)\right]$ at the point $x=0$

where n is even.

Answer 1

$f\left(x\right)=\frac{1}{1-x^2}=\frac{1}{(1-x)(1+x)}=\frac{1}{2}[\frac{1}{1-x}+\frac{1}{1+x}]$

$\frac{d^{n}f\left(x\right)}{d{x}^n}=\frac{1}{2}\frac{d^n}{d{x}^n}\left(\frac{1}{1+x}\right)+\frac{1}{2}\frac{d^n}{d{x}^n}\left(\frac{1}{1+x}\right)$

$=\frac{1}{2}\frac{{(-1)}^n.n!{(-1)}^n}{{(1-x)}^{n+1}}+\frac{1}{2}\frac{{(-1)}^n.n!{\left(1\right)}^n}{{(1+x)}^{n+1}}$

When $n$ is even

$=\frac{1}{2}\frac{n!}{{(1-x)}^{n+1}}+\frac{1}{2}\frac{n!}{{(1+x)}^{n+1}}$

$\therefore$ at $x=0$

$\frac{d^{n}f\left(x\right)}{d{x}^n}=\frac{1}{2}.\frac{n1}{1}+\frac{1}{2}.\frac{n!}{1}$

$=\frac{1}{2}.2n!$

$\frac{d^{n}f\left(o\right)}{d{x}^n}=n!$ , when $n$ is even .