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Let $a$ be a constant. Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^2-a^2}$.

After computing a few derivatives, the derivatives seem to have factorials in them sometimes and other times not. For example, $\dfrac{d^5}{dx^5} \left (\dfrac{1}{x^2-a^2} \right) = -\dfrac{240(3a^4x+10a^2x^3+3x^5)}{(x^2-a^2)^6}$ while $\dfrac{d^6}{dx^6} \left (\dfrac{1}{x^2-a^2} \right) = -\dfrac{720(a^6+21a^4x^2+35a^2x^4+7x^6)}{(x^2-a^2)^6}$, so I am not immediately seeing the pattern.

user19405892
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2 Answers2

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Just observe that $$f(x)=\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right)$$

Then $$f^{(n)}(x)=\frac{(-1)^nn!}{2a}\left[\frac{1}{(x-a)^{n+1}}-\frac{1}{(x+a)^{n+1}}\right]$$

Ángel Mario Gallegos
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    If should be $n+1$ in the exponent in the denominators: $\frac{1}{(x-a)^{n+1}}$. – Thomas Andrews Mar 16 '16 at 23:36
  • And the coefficients of the numerator are seen to be binomial coefficients. You get $$f^{n}(x)=(-1)^nn!\frac{\sum_{k\text{ odd}}\binom{n}{k}x^{n-k}a^{k-1}}{(x^2-a^2)^{n+1}}$$ – Thomas Andrews Mar 16 '16 at 23:39
  • @ThomasAndrews How did you simplify it like that? – user19405892 Mar 17 '16 at 01:32
  • Actually, my formula was wrong. But only because I forgot my previous comment. The actual closed form for the numerator is:$$\sum_{k\text{ odd}} \binom{n+1}{k}x^{n+1-k}x^{n-k}a^{k-1}$$. This is the value of $$\frac{1}{2a}\left((x+a)^{n+1}-(x-a)^{n+1}\right)$$. The even terms cancel, so you are left with the odd terms. – Thomas Andrews Mar 17 '16 at 01:44
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$$ {1 \over {x^{\,2} - a^{\,2} }} = {1 \over {\left( {x + a} \right)\left( {x - a} \right)}} = {1 \over {2\,a}}\left( {{1 \over {\left( {x - a} \right)}} - {1 \over {\left( {x + a} \right)}}} \right) $$ ... then it goes easily ..

G Cab
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