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Can anyone help me finding the $n^{th}$ derivative with respect to $x$ of the function

$$\frac{f(x)}{x-a}$$ where $f$ is infinitely differentiable, $a$ is some constant. I tried to find the first few terms but things get messy!

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4 Answers 4

As many have noted, Leibniz's rule is applicable here. First, we note that

$$(f \cdot g)^{(n)}=\sum_{k=0}^n {n \choose k} f^{(k)} g^{(n-k)}.$$

Further,

$$\left(\frac{1}{(x-a)}\right)^\prime=-1\cdot\frac{1}{(x-a)^2}$$

and

$$\left(\frac{1}{(x-a)}\right)^{\prime\prime}=-1\cdot-2\cdot\frac{1}{(x-a)^3}.$$

In general,

$$\left(\frac{1}{x-a}\right)^{(n)}=(-1)^n\frac{n!}{(x-a)^{n+1}}.$$

Using this information, we see that

$$\left(\frac{f(x)}{x-a}\right)^{(n)}=\sum_{k=0}^n {n \choose k} f^{(k)}(x) (-1)^{n-k}(n-k)!\frac{1}{(x-a)^{n-k+1}}.$$

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Hint. $$\begin{align*} (fg)' & = f'g + fg'\\ (fg)'' &= f''g + 2f'g' + fg''\\ (fg)''' &= f'''g + 3f''g' + 3f'g'' + fg'''\\ (fg)^{(4)} &= f^{(4)}g + 4f^{(3)}g' + 6f^{(2)}g^{(2)} + 4f'g^{(3)} + fg^{(4)} \end{align*}$$ See the pattern?

Prove it by induction, then apply it with $g(x) = (x-a)^{-1}$.

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Use the Leibniz rule.

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in general $$(fg)^{(n)}=\sum_{k=0}^n {n\choose k} f^{(k)}g^{(n-k)}$$

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