Finding the nth derivative of a function

[TheRainMan713]

Homework Statement

I'm trying to find a formula for the nth derivative for the function f(x)=x^1/3

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...

 

[VrhoZna]

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)x^1/3-n

 

[Ray Vickson]

Homework Statement

I'm trying to find a formula for the nth derivative for the function f(x)=x^1/3

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.