# Finding the nth derivative of a function

## Homework Statement

I'm trying to find a formula for the nth derivative for the function f(x)=x1/3

## The Attempt at a Solution

I know that it has alternating signs so it start with (-1)n+1 and I know the exponent for it is x(1/3-n) but I'm having a hard time figuring out the coefficient of x.

For the fourth derivative I have 1/3(1/3-1)(1/3-2)(1/3-3)x(1/3-4)

The example our teacher gave us was x-1 which was much easier in my opinion...

Try expressing the powers and the coefficients in a fractional form while taking derivatives and see if the pattern becomes clearer that way.

TheRainMan713
I rewrote it three different ways and I'm still having a hard time seeing the complete pattern.

I'm using the fourth derivative: f4(x)=(1/3)(-2/3)(-5/3)(-8/3)x-11/3
and: f4(x)=(1/3)(1/3-1)(1/3-2)(1/3-3)x-11/3
and:f4(x)=(1/3)(1/3-3/3)(1/3-6/3)(1/3-9/3)x-11/3

So far I have fn(x)=(-1)n+1(?/3n)x1/3-n

Ray Vickson
Homework Helper
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## Homework Statement

I'm trying to find a formula for the nth derivative for the function f(x)=x1/3

## The Attempt at a Solution

I know that it has alternating signs so it start with (-1)n+1 and I know the exponent for it is x(1/3-n) but I'm having a hard time figuring out the coefficient of x.

For the fourth derivative I have 1/3(1/3-1)(1/3-2)(1/3-3)x(1/3-4)

The example our teacher gave us was x-1 which was much easier in my opinion...

IMHO it is much easier to work with the general form ##f(x) = x^p##, then take ##p = 1/3## after all the work is finished.

The reason it is easier is that you have symbolic factors such as ##p - 1##, ##p-2,## etc. and by keeping them symbolic you can keep straight the different "effects". For example, if you see a number like ##-2## somewhere in your calculation, it is not easy to know if it really is a ##-2"## or a ##-1 - 3/3"## or ##-6/3"##---and sometimes that matters a lot when you want to look at more terms, etc. By keeping everything in terms of ##p## there is never any chance of confusion.