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

How can I get a general formula for the nth derivative of $f(x)^k$ with respect to x, in terms of other derivatives of $f(x)$?

In other words, I need a general formula for $$\frac{d^n}{dx^n}f(x)^k$$ Where k is a fixed integer, I would appreciate any help

Note: The OP originally asked about: $$\frac{d^n}{dx^n}\ln(f(x))$$ Some of the comments or solutions below may refer to the original question.

share|cite|improve this question
It won't be nice. In fact, it'll probably be uselessly complicated. What do you need it for? – icurays1 Jan 29 '13 at 17:20
Its fine if it is complicated, just try to condense it as much as posible if you can, thanks. – Ethan Jan 29 '13 at 17:25
did you seriously change your question from $\frac{d^n}{dx^n}\ln(f(x))$ to $\frac{d^n}{dx^n}f(x)^k$? Sure this is the last edit of this magnitude? -.- – example Jan 29 '13 at 23:55

Probably the only halfway nice formula would come from using the iterated product rule:

$$ \frac{d^n}{dx^n}(uv)=\sum_{k=0}^n {n\choose k}u^{(n-k)}v^{(k)} $$on the function

$$ \frac{d}{dx}\ln(f(x))=\frac{f^\prime(x)}{f(x)}. $$ You would use say $u=f^\prime$ and $v=(f(x))^{-1}$. Of course, the problem is still nasty because now you need an expression for the $n$th derivative of $(f(x))^{-1}$ (more iterated product rule with some chain rule).

share|cite|improve this answer

It has the form of $P/f^n(x)$ where $P$ is a degree $n$ polynomial of $f(x), f'(x), ... , f^{(n)} (x)$.

share|cite|improve this answer

This is not pretty at all. If you know that $f(x)$ is sufficiently close to $x$, Perhaps you can use: $$\frac{\partial^n log(x)}{\partial x^n} =\frac{(-1)^n(n-1)!}{x^n}$$ And then try to plug in $x\rightarrow f(x)$.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.