Question

lf $y=\frac{1}{x^2-a^2}$ then $y_n=$
(where $y_n$ denotes the $n^{th}$ derivative of $y$ w.r.t. $x$)

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

Answer 1

Answer: (B)

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

$y=\frac{1}{(x-a)(x+a)}$
$=\frac{1}{2a}(\frac{1}{x-a}-\frac{1}{x+a})$
$\frac{dy}{dx}=\frac{1}{2a}(\frac{-1}{(x-a{)}^2}-(-\frac{1}{(x+a{)}^2}))$
$\frac{d^{2}y}{d{x}^2}=\frac{1}{2a}(\frac{2!}{(x-a{)}^3}-\left(\frac{2!}{(x+a{)}^3}\right))$
$\frac{d^{3}y}{d{x}^3}=\frac{-1}{2a}(\frac{3!}{(x-a{)}^4}-\left(\frac{3!}{(x+a{)}^4}\right))$
Following the pattern we get
$\frac{d^{n}y}{d{x}^n}=\frac{(-1{)}^n}{2a}n!(\frac{1}{(x-a{)}^{n+1}}-\frac{1}{(x+a{)}^{n+1}})$