# Taylor series integration

1. Oct 15, 2008

### okkvlt

Questions:

Is there a quicker way to find the formula for the nth derivative of a function, instead of finding the first several derivatives and trying to find a pattern, and using that pattern to form the equation for the nth derivative?

Also, is there a formula for the nth derivative of x^x? I cant find a formula for its derivatives so i cant find the polynomial. By the third derivative it starts to get complicated, and by the 5th i need to use 4 lines on a sheet of notebook paper just to write the derivative. Then i gave up because it got too cumbersome. when x^x is diffferentiated, it gives a x^x(1+lnx), and when that's differentiated, it gives a x^(x-1), then that gives a x^(x-1)*(1+lnx-1/x) Those terms keep producing more and more things, all the while the x^x(1+lnx)^n and x^(x-1)*(1+lnx-1/x)^n keep adding more and more things to the equation until it becomes a monster.

maybe if i tried differentiating logx(y)=x, but i cant figure out how to do that in the first place. I have a feeling that it will be even more complicated, because implicit differentation would be required.

Also im curious, is x^x a function one step higher than exponentation? Just as exponentation is a step higher than multiplication and multiplication is a step higher than addition.

And, how do i integrate the series of a series? like if the x of a taylor polynomial is another taylor polynomial. If that integration is impossible, then how do i find the taylor polynomial that represents a taylor polynomial of a taylor polynomial?

2. Oct 17, 2008

### Gib Z

Not in any "useful" form. Of course, $$\frac{d^n}{dx^n}$$ does it quite well.