Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^3-a^2x}$

Asked Mar 17 '16 at 03:10

Active Mar 17 '16 at 03:10

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Let $a$ be a constant. Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^3-a^2x}$.

I think if we separated this two fractions it might help this result. We then have $\dfrac{1}{x^3-a^2x} = \dfrac{1}{x} \cdot\dfrac{1}{x^2-a^2}$. Is there way to use the value of the second terms $n$th derivative to find the $n$th derivative of the $f(x)$ we want?

edited Apr 13 '17 at 12:19

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asked Mar 17 '16 at 03:10

user19405892

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Have you tried decomposing into simple fractions? – Catalin Zara Mar 17 '16 at 03:13
You can do partial fraction decomposition--then the derivatives should all be "nice" (I think). Since that can be written as $f(x) = \frac{1}{x(x - a)(x + a)}$ – Jared Mar 17 '16 at 03:13
Ah, that does it. Thanks! – user19405892 Mar 17 '16 at 03:18
Here is a related problem. – Mhenni Benghorbal Mar 17 '16 at 03:36

Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^3-a^2x}$

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