Question

If $y={(x^2-1)}^m$, then the ${\left(2m\right)}^{th}$ differential coefficient of $y$ with respect to $x$ is:

• A. $m$
• B. $\left(2m\right)!$
• C. $m!$
• D. $2m$

Answer 1

The correct option is B

$\left(2m\right)!$

Explanation for the correct option.

Finding the differential coefficient of $y$:

${\left(2m\right)}^{th}$ differential coefficient of $y$ with respect to $x$ means, we have to find $\frac{d^{2m}y}{d{x}^{2m}}$.

$\begin{array}{rcl}y& =& {(x^2-1)}^m={}^{m}C_0{\left(x^2\right)}^m{\left(1\right)}^0-{}^{m}C_1{\left(x^2\right)}^{m-1}{\left(1\right)}^1-{}^{m}C_2{\left(x^2\right)}^{m-2}{\left(1\right)}^2..........\\ & =& {}^{m}C_0\left(x^{2m}\right)-{}^{m}C_1\left(x^{2m-2}\right)-{}^{m}C_2\left(x^{2m-4}\right).......\end{array}$

Here, the highest power is $2m$, and we have to find $\frac{d^{2m}y}{d{x}^{2m}}$, so we will consider $\begin{array}{rcl}& & {}^{m}C_0\left(x^{2m}\right)\end{array}$ only.

Now, the first derivative will be:

$\frac{dy}{dx}={}^{m}C_0\left(2m\right)\begin{array}{l}\left(x^{2m-1}\right)\end{array}$

The second derivative will be:

$\frac{d^{2}y}{d{x}^2}={}^{m}C_0\left(2m\right)\left(2m-1\right)\left(x^{2m-2}\right)$

Similarly, the ${\left(2m\right)}^{th}$ derivative will be

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& {}^{m}C_0\left(2m\right)\left(2m-1\right).........\left\{2m-\left(2m-1\right)\right\}\left(x^0\right)\\ & =& \left(2m\right)\left(2m-1\right)\left(2m-2\right).........1\\ & =& \left(2m\right)!\end{array}$

Hence, option B is correct.