Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question

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

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





In general,


Using this information, we see that

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

share|cite|improve this answer

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 $g(x) = (x-a)^{-1}$.

share|cite|improve this answer

Use the Leibniz rule.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.