Question

How do you find the nth derivative of $\dfrac{1}{{{x}^{2}}-9}$ ?

Answer 1

To find the nth derivative of a given function we don’t have a general formula, so we have to follow a few steps in order to find the nth derivative of a function. First, we will differentiate the given function to find its 1st order 2nd order and 3rd order derivatives. After that we can see some changes of power of the function, additional coefficient and many more and then use these changes to express it in nth derivative. This case was for standard functions but in case of non-standard functions we first express them in standard function then we use the general formula of nth derivative of a standard function in order to find the nth derivative of a non-standard function.

Complete answer:
In the above question we have the nth derivative of $\dfrac{1}{{{x}^{2}}-9}$ . This expression can be written as
$\dfrac{1}{{{x}^{2}}-9}=\dfrac{1}{{{x}^{2}}-{{3}^{2}}}=\left[ \dfrac{x+3}{x-3}-\dfrac{x-3}{x+3} \right]=\dfrac{1}{6}\left[ \dfrac{1}{x-3}-\dfrac{1}{x+3} \right]$
So we can say that $f(x)=\dfrac{1}{(x+a)}$
Now we will find the 1st order 2nd order and 3rd order derivatives.

& \dfrac{d}{dx}f(x)=-\dfrac{1}{{{(x+a)}^{2}}} \\\ & \dfrac{{{d}^{2}}}{d{{x}^{2}}}f(x)=-\dfrac{(-1)(-2)}{{{(x+a)}^{3}}} \\\ & \dfrac{{{d}^{3}}}{d{{x}^{3}}}f(x)=-\dfrac{(-1)(-2)(-3)}{{{(x+a)}^{4}}} \\\ \end{aligned}$$ From the observation we can say that the general formula for the above function is $$\dfrac{{{d}^{n}}}{d{{x}^{n}}}f(x)=-\dfrac{{{(-1)}^{n}}n!}{{{(x+a)}^{n+1}}}$$ Now in place of $$f(x)$$ we will substitute the given function in the question. $$\begin{aligned} & \dfrac{{{d}^{n}}}{d{{x}^{n}}}\left( \dfrac{1}{{{x}^{2}}-9} \right)=\dfrac{{{d}^{n}}}{d{{x}^{n}}}\left( \dfrac{1}{6}\left[ \dfrac{1}{x-3}-\dfrac{1}{x+3} \right] \right) \\\ & \dfrac{{{d}^{n}}}{d{{x}^{n}}}\left( \dfrac{1}{{{x}^{2}}-9} \right)=\left( \dfrac{1}{6}\left[ \dfrac{{{d}^{n}}}{d{{x}^{n}}}\left( \dfrac{1}{x-3} \right)-\dfrac{{{d}^{n}}}{d{{x}^{n}}}\left( \dfrac{1}{x+3} \right) \right] \right) \\\ & \dfrac{{{d}^{n}}}{d{{x}^{n}}}\left( \dfrac{1}{{{x}^{2}}-9} \right)=\dfrac{1}{6}\left[ \dfrac{{{(-1)}^{n}}n!}{{{(x-3)}^{n+1}}}-\dfrac{{{(-1)}^{n}}n!}{{{(x+3)}^{n+1}}} \right] \\\ & \dfrac{{{d}^{n}}}{d{{x}^{n}}}\left( \dfrac{1}{{{x}^{2}}-9} \right)=\dfrac{{{(-1)}^{n}}n!}{6}\left[ {{(x-3)}^{n+1}}-{{(x+3)}^{n+1}} \right] \\\ \end{aligned}$$ Thus, we can conclude that the nth derivative for $$\left( \dfrac{1}{{{x}^{2}}-9} \right)$$ is $$\dfrac{{{(-1)}^{n}}n!}{6}\left[ {{(x-3)}^{n+1}}-{{(x+3)}^{n+1}} \right]$$ . **Note:** This question can also be solved with the help of Leibnitz rule. Since, this question involves a lot of calculation so solve it carefully in order to avoid silly mistakes. Keep in mind the procedure used to solve the question for future use. Solve the question step by step to avoid confusion. Correctly derive the general formula.