Question

For $4y=\frac{x}{x^2-1},\frac{d^{n}y}{d{x}^n}$ is equal to.

• A. $\frac{n!}{2}\left[\frac{1}{(x-1{)}^n}+\frac{1}{(x-1{)}^n}\right]$
• B. $\frac{(-1{)}^{n}n!}{2}\left[\frac{1}{(x+1{)}^n}-\frac{1}{(x-1{)}^n}\right]$
• C. $\frac{n!}{2}\left[\frac{1}{(x+1{)}^{n+1}}-\frac{1}{(x-1{)}^{n+1}}\right]$
• D. $\frac{(-1{)}^{n}n!}{2}\left[\frac{1}{x+1{)}^{n+1}}-\frac{1}{x-1{)}^{n+1}}\right]$

Answer 1

Answer: (D)

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

We can write $\frac{x}{x^2-1}$ using partial fraction: $\frac{x}{x^2-1}=\frac{1}{2}[\frac{1}{x+1}+\frac{1}{x-1}]$

$4\frac{dy}{dx}=\frac{1}{2}[\frac{-1}{{(x+1)}^2}+\frac{-1}{{(x-1)}^2}]$

$4\frac{d^{2}y}{d{x}^2}=\frac{1}{2}[+\frac{2(x+1)}{{(x+1)}^4}+\frac{2(x-1)}{{(x-1)}^4}]$

$=2\times \frac{1}{2}[2\frac{1}{{(x+1)}^3}+\frac{2}{{(x-1)}^3}]$

$4\frac{d^{3}y}{{dx}^2}=\frac{2}{2}[-\frac{3{(x+1)}^2}{{(x+1)}^6}+\frac{-3{(x-1)}^2}{{(x-1)}^6}]$

$=\frac{3\times 2}{2}[-\frac{1}{{(x+1)}^4}+\frac{-1}{{(x+1)}^4}]$

Hence,
$4\frac{d^{n}y}{{dx}^n}=\frac{{(-1)}^{n}n!}{2!}[\frac{1}{{(1+x)}^{n+1}}+\frac{1}{{(x-1)}^{n+1}}]$