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

Hint: 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.