# Calculus Derivative help

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.

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It won't be nice. In fact, it'll probably be uselessly complicated. What do you need it for? –  icurays1 Jan 29 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 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 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).

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