# Nth Derivative

1. Nov 20, 2005

### daniel_i_l

Is there a way to find the nth derivative of a function (at least for polynomials) or does a knew trick need to be madeup for every one?

2. Nov 20, 2005

### Hypnos

Don't you just apply the same rules repeatedly when taking derivatives? What tricks are you referring to?

3. Nov 20, 2005

### BerkMath

There are several tricks when calculating nth derivatives of functions. The more interesting case is in complex analysis when one may make use of Cauchy's integral formula for the nth derivative. Its quite fascinating that the value of the nth derivative of a function within a contour can be found only by evaluating the functions values on the contour. If you are interested in a specific class of functions then we could tell you exactly what is at your disposal.

4. Nov 20, 2005

### daniel_i_l

mostly complicated polynomials.

5. Nov 21, 2005

### HallsofIvy

Staff Emeritus
I find the very idea that a polynomial could be "complicated" puzzling!

You can differentiate any polynomial by using the power rule. Each derivative is a polynomial of degree one lower than the original so it should become easier to find higher derivatives. Perhaps I am misunderstanding. Other than the power rule you don't need any "tricks".

If by "complicated" you mean written as a product of terms, either go ahead and multiply them out or use the product rule and chain rule.

6. Nov 24, 2005

### daniel_i_l

I meant polynomials with more than one term, and I don't just wan't to differentiate them, I wan't to find the nth derivative. For example, the nth derivative of a^n is n! .

7. Nov 24, 2005

### mathphys

Ok, if i understand what you are trying to do

then you are asking yourself what is the kth derivative of x^n (a 'general expresion for it')?

Well, think in the falling factorial symbol instead of the simple factorial and you will get the answer.

For a polynomial a0+a1x+a2x^2+...+ajx^j+...+anx^n you just need to apply the correspoinding rule to each term, with the condition than if k>j the kth derivative of that term vanishes.

8. Nov 29, 2005

### Masoud

Teh interesting thing is that I foudn this place because I was trying to see if a formula existed. I have derived a very simple formula for teh nth derivative just today. It's quite easy. Write them out for $y=ax^n$, and you'll see a pattern...the rest is easy to formulate. :)
Masoud Zargar