Question

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

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

Answer 1

The correct option is A

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

The explanation for the correct option:

Step 1. Modification:

We have $y=\frac{x}{(x^2-1)}$

$$\begin{gathered} =\frac{x}{(x-1)(x+1)} \\ =\frac{2x}{2(x-1)(x+1)} \\ =\left(\frac{1}{2}\right)\left[\frac{(x+1)+(x-1)}{(x-1)(x+1)}\right] \\ =\left(\frac{1}{2}\right)\left[\frac{1}{(x-1)}+\frac{1}{(x+1)}\right] \end{gathered}$$

Step 2. Differentiation with respect to $x$ :

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

Again differentiation with respect to $x$, we get

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\left(\frac{1}{2}\right)\left[\frac{2}{{(x-1)}^3}+\frac{2}{{(x+1)}^3}\right]$

Now, differentiation with respect to $x$ n times, we get

$$\begin{gathered} \Rightarrow \frac{d^{n}y}{d{x}^n}=\left(\frac{1}{2}\right)\left[\frac{{(-1)}^{n}n!}{{(x-1)}^{n+1}}-\frac{{(-1)}^{n}n!}{{(x+1)}^{n+1}}\right] \\ \Rightarrow \frac{d^{n}y}{d{x}^n}=\left(\frac{1}{2}\right){(-1)}^{n}n!\left[\frac{1}{{(x-1)}^{n+1}}+\frac{1}{{(x+1)}^{n+1}}\right] \end{gathered}$$)

Hence, the correct option is (A).