Question

If we have an expression as \[y = {\left( {{x^2} - 1} \right)^m}\], then the \[{\left( {2m} \right)^{th}}\] differential coefficient of \[y\] w.r.t. \[x\] is
(A) \[m\]
(B) \[\left( {2m} \right)!\]
(C) \[2m\]
(D) \[m!\]

Answer 1

Hint: To find the \[{\left( {2m} \right)^{th}}\] differential coefficient of \[y\] w.r.t. \[x\], we use the Binomial theorem to expand \[{\left( {{x^2} - 1} \right)^m}\]. We find that \[2m\] is the highest power of expansion. Therefore, we need to consider the term containing \[{x^{2m}}\] only. Now differentiate the expression \[2m\] times with respect to \[x\] to find the result.

Complete step-by-step solution:
From Binomial theorem, we know that, if \[x\] and \[a\] are real numbers, then for all \[n \in N\],
\[{\left( {x + a} \right)^n}{ = ^n}{C_0}{x^n}{a^0}{ + ^n}{C_1}{x^{n - 1}}{a^1}{ + ^n}{C_2}{x^{n - 2}}{a^2} + ...{ + ^n}{C_r}{x^{n - r}}{a^r} + ...{ + ^n}{C_{n - 1}}{x^1}{a^{n - 1}}{ + ^n}{C_n}{x^0}{a^n}\]
i.e., \[{\left( {x + a} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{a^r}} \]
Given \[y = {\left( {{x^2} - 1} \right)^m}\]
Expanding \[{\left( {{x^2} - 1} \right)^m}\] using binomial theorem,
\[{\left( {{x^2} - 1} \right)^m}{ = ^m}{C_0}{\left( {{x^2}} \right)^m}{\left( { - 1} \right)^0}{ + ^m}{C_1}{\left( {{x^2}} \right)^{m - 1}}{\left( { - 1} \right)^1}{ + ^m}{C_2}{\left( {{x^2}} \right)^{m - 2}}{\left( { - 1} \right)^2} + ...\]
On simplifying,
\[{\left( {{x^2} - 1} \right)^m}{ = ^m}{C_0}{\left( {{x^2}} \right)^m}{ - ^m}{C_1}{\left( {{x^2}} \right)^{m - 1}}{ + ^m}{C_2}{\left( {{x^2}} \right)^{m - 2}} - ...\]
On further simplification,
\[{\left( {{x^2} - 1} \right)^m}{ = ^m}{C_0}{x^{2m}}{ - ^m}{C_1}{x^{2m - 2}}{ + ^m}{C_2}{x^{2m - 4}} - ...\]
Here the highest power of x is \[2m\].
So on differentiating \[2m\] times, all other terms will become zero.
Therefore, to find the \[{\left( {2m} \right)^{th}}\] differential coefficient we will only consider the term \[^m{C_0}{x^{2m}}\].
Therefore, we get \[y{ = ^m}{C_0}{x^{2m}} - - - (1)\].
We know, \[^m{C_0}\] is a constant.
On differentiating both the sides of \[(1)\] with respect to \[x\], we get
\[ \Rightarrow \dfrac{{dy}}{{dx}}{ = ^m}{C_0}\left( {2m} \right){x^{2m - 1}}\]
On again differentiating, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}}{ = ^m}{C_0}\left( {2m} \right)\left( {2m - 1} \right){x^{2m - 2}}\]
Similarly, on differentiating \[2m\] times, we get
\[ \Rightarrow \dfrac{{{d^{2m}}y}}{{d{x^{2m}}}}{ = ^m}{C_0}\left( {2m} \right)\left( {2m - 1} \right)...\left( {2m - \left( {2m - 1} \right)} \right){x^0} - - - (2)\]
As we know,
\[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\]
On simplifying \[\left( 2 \right)\] using this, we get
\[ \Rightarrow \dfrac{{{d^{2m}}y}}{{d{x^{2m}}}} = \dfrac{{m!}}{{\left( {m - 0} \right)!0!}}\left( {2m} \right)\left( {2m - 1} \right)...1\]
As \[0! = 1\], we get
\[ \Rightarrow \dfrac{{{d^{2m}}y}}{{d{x^{2m}}}} = \left( {2m} \right)\left( {2m - 1} \right)...1\]
On simplifying, we get
\[ \Rightarrow \dfrac{{{d^{2m}}y}}{{d{x^{2m}}}} = \left( {2m} \right)!\]
Therefore, the \[{\left( {2m} \right)^{th}}\] differential coefficient of \[y\] w.r.t. \[x\] of \[y = {\left( {{x^2} - 1} \right)^m}\] is \[\left( {2m} \right)!\].
Hence, option (B) is correct.

Note: Here the highest power of \[x\] is \[2m\] and also, we have to find the \[{\left( {2m} \right)^{th}}\] differential coefficient of \[y\] w.r.t. \[x\], for this type of problems we get a constant value of differential coefficient. In this problem all terms having power of \[x\] less than \[2m\] on differentiating will become zero so there is no need to consider those terms as it will make the question complicated. For this type of problems consider only the highest degree term and differentiate it accordingly to get the result.