# Taylor expansion of x^x

1. Sep 17, 2010

### danik_ejik

hello,
[URL]http://latex.codecogs.com/gif.latex?f(x)=x^{x}-1[/URL] around the point a=1

i thought to write it as g(x)=x^x
and then f(x)=g(x)-1
and then find the polynomial for g(x) as lng(x)=xln(x)
but it seems incorrect.

Last edited by a moderator: Apr 25, 2017
2. Sep 17, 2010

### hunt_mat

In order to do this you have to calculate the derivative if $$y=x^{x}$$, take logs of this equation and differentiate that using implicit differentiation and that will help you, or you could write:
$$x^{x}=e^{x\log x}$$
And use the chain rule

3. Sep 17, 2010

### danik_ejik

thanks,
successfully managed by directly calculating the taylor polynomial when
[URL]http://latex.codecogs.com/gif.latex?f(x)=e^{xlnx}[/URL]

Last edited by a moderator: Apr 25, 2017