# [For experts] Derivatives of 1/f(x)^2

1. Sep 28, 2005

### wackensack

My question is presented in the uploaded pdf file.

:surprised

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2. Sep 29, 2005

### loloPF

But You might want to know that:
$$\frac{d}{dx}(\frac{1}{f^2(x)})=-2\frac{f'(x)}{f^3(x)}$$

3. Sep 29, 2005

### Jameson

I just briefly looked at the problem, but I wanted to say that the file is fine. Just a math problem. :)

4. Sep 29, 2005

### Hurkyl

Staff Emeritus
I have a nitpick -- smooth functions (i.e. infinitely differentiable) are not required to have a MacLauren series -- you need to be analytic.

5. Oct 5, 2005

### benorin

Use Faá di Bruno's Formula for the Nth derivative of a composition of functions (since 1/f(x)^2=h(f(x)) where h(x)=1/x^2). Here is a link:

hope you like integer partitions...

6. Oct 9, 2005

### wackensack

It may help

Thank you, Mr. Benorin. I'm trying to adapt the Faá di Bruno's formula to my problem.

Bob

7. Jan 23, 2006

### benorin

Ok, so I found another formulation of Faa di Bruno's formula for the nth derivative of a composition of functions: here's your answer

$$\frac{d^{n}}{dx^{n}}\left(-\frac{1}{f^{2}(x)}\right) = \sum_{m=1}^{n}\left\{\frac{1}{m!}\left[\sum_{j=0}^{m-1}(-1)^{j}\frac{m!}{j!(m-j)!}f^{j}(x)\frac{d^{n}}{dx^{n}}\left( f^{m-j}(x)\right)\right]\frac{(-1)^{m+1}(m+1)!}{f^{m+2}(x)}\right\}$$

where $$f^{k}(x)$$ is the kth power of f(x) (not the kth derivative.)

-Ben

8. Jan 28, 2006

### wackensack

Mr. Benorin, you see, this is a local problem: the final result is evaluated at $$x = a$$. Besides that, $$f$$ satisfies some particular conditions, which must be considered:
(a) $$f(x) \neq 0$$, over some open interval $$A$$;
(b) $$f$$ is a series of even powers;
(c) $$f^{(2n+1)}(a) = 0$$ and $$f^{(2n)}(a) \neq 0$$, $$n = 0, 1, 2, ...$$;
The final result is a function of $$a$$, and the sum symbol, $$\Sigma$$, will not appear in the final answer.
As I've pointed,
$$g^{(2n)}(a)=-b_n f(a)^{-3n-2}$$​
Find $$(b_n)$$​
$$(b_n) = (1, 22, 584, 28384, 2190128, ...)$$​