Question

If $y = \dfrac{1}{{{x^2} - {a^2}}}$ then ${y_n} =$
A. $\dfrac{{{{\left( { - 1} \right)}^n}n!}}{{2a}}\left[ {\dfrac{1}{{{{\left( {x - a} \right)}^n}}} - \dfrac{1}{{{{\left( {x + a} \right)}^n}}}} \right]$
B. $\dfrac{{{{\left( { - 1} \right)}^n}n!}}{{2a}}\left[ {\dfrac{1}{{{{\left( {x - a} \right)}^{n + 1}}}} - \dfrac{1}{{{{\left( {x + a} \right)}^{^{n + 1}}}}}} \right]$
C. $\dfrac{{{{\left( { - 1} \right)}^n}n!}}{{2a}}\left[ {\dfrac{1}{{{{\left( {x - a} \right)}^{n + 1}}}} + \dfrac{1}{{{{\left( {x + a} \right)}^{^{n + 1}}}}}} \right]$
D. $\dfrac{{{{\left( { - 1} \right)}^n}n!}}{{2a}}\left[ {\dfrac{1}{{{{\left( {x - a} \right)}^n}}} + \dfrac{1}{{{{\left( {x + a} \right)}^n}}}} \right]$

Answer 1

In the above given question, we are given an equation $y = \dfrac{1}{{{x^2} - {a^2}}}$ . We have to find the nth differentiation of $y$ i.e. we have to find the value of ${y_n}$ . In order to approach the solution, we need to find the few consecutive initial derivatives of $y$ . We can use the method of partial fraction in order to calculate the derivative of the equation $y = \dfrac{1}{{{x^2} - {a^2}}}$.

Complete step by step answer:
Given that, the equation $y = \dfrac{1}{{{x^2} - {a^2}}}$. We have to find the nth derivative of the above given equation i.e. we have to find ${y_n}$. Since the given equation is of the form $y = \dfrac{1}{{{x^2} - {a^2}}}$ , hence we can also write it in the form
$\Rightarrow y = \dfrac{1}{{\left( {x - a} \right)\left( {x + a} \right)}}$
Now we can split the denominator in two parts using the method of partial fraction, such as
$\Rightarrow y = \dfrac{1}{{2a}}\left[ {\dfrac{1}{{\left( {x - a} \right)}} - \dfrac{1}{{\left( {x + a} \right)}}} \right]$

Now differentiating above equation with respect to $x$ , we get
$\Rightarrow y' = \dfrac{1}{{2a}}\left[ {\dfrac{{ - 1}}{{{{\left( {x - a} \right)}^2}}} - \dfrac{{\left( { - 1} \right)}}{{\left( {x + a} \right)}}} \right]$
Similarly, we have to obtain a few more consecutive derivatives of $y$ .
Therefore, differentiating again gives us,
$\Rightarrow y'' = \dfrac{1}{{2a}}\left[ {\dfrac{{2!}}{{{{\left( {x - a} \right)}^3}}} - \dfrac{{2!}}{{{{\left( {x + a} \right)}^3}}}} \right]$
Again, differentiating the above equation gives us,
$\Rightarrow y''' = \dfrac{{ - 1}}{{2a}}\left[ {\dfrac{{3!}}{{{{\left( {x - a} \right)}^4}}} - \dfrac{{3!}}{{{{\left( {x + a} \right)}^4}}}} \right]$
And once again, differentiating the above equation gives us,
$\Rightarrow {y_{^{\left( 4 \right)}}} = \dfrac{1}{{2a}}\left[ {\dfrac{{4!}}{{{{\left( {x - a} \right)}^5}}} - \dfrac{{4!}}{{{{\left( {x + a} \right)}^5}}}} \right]$

And so on. Using this pattern, many more further derivatives of $y$ can be obtained.
Therefore, if we notice the pattern of each differentiation of $y$ , then we can identify a unique pattern of writing each term. Following the above pattern, we can write the nth derivative of the equation $y = \dfrac{1}{{{x^2} - {a^2}}}$ as,
$\Rightarrow {y_{^n}} = \dfrac{{{{\left( { - 1} \right)}^n}}}{{2a}}\left[ {\dfrac{{n!}}{{{{\left( {x - a} \right)}^{n + 1}}}} - \dfrac{{n!}}{{{{\left( {x + a} \right)}^{n + 1}}}}} \right]$
That gives us,
$\Rightarrow {y_{^n}} = \dfrac{{{{\left( { - 1} \right)}^n}}}{{2a}}n!\left[ {\dfrac{1}{{{{\left( {x - a} \right)}^{n + 1}}}} - \dfrac{1}{{{{\left( {x + a} \right)}^{n + 1}}}}} \right]$
That is the required nth derivative of $y$ .
Therefore, if $y = \dfrac{1}{{{x^2} - {a^2}}}$ then ${y_{^n}} = \dfrac{{{{\left( { - 1} \right)}^n}}}{{2a}}n!\left[ {\dfrac{1}{{{{\left( {x - a} \right)}^{n + 1}}}} - \dfrac{1}{{{{\left( {x + a} \right)}^{n + 1}}}}} \right]$ .

Hence, the correct option is B.

Note: There is no general formula to find the nth derivative of a function. Finding the nth derivative means to take a few derivatives (1st, 2nd, 3rd and so on) of that function and look for a pattern. If that exists, then we have a formula for its nth derivative. In order to find the nth derivative, find the first few derivatives to identify the pattern. Apply the usual rules of differentiation to a function. Find each successive derivative to arrive at the nth derivative of the function.