Finding second and third derivative for Taylor series

I'm calculating the 3rd degree Taylor polynomial for $$y =$$ $$x^{x^x}$$ in $$x_0=1$$. I calculated the first derivative, which is $$e^{lnx*x^x}*(\frac{x^x}{x}+lnx*e^{lnx*x}*(1+lnx))$$. Finding the first degree derivative in $$x_0=1$$ is very easy, but calculating the second and third derivative is not.

Is there any way I can simplify it or do I really have to find all the lengthy second and third degree derivatives of $$y$$?

Thanks

• $x_0=0$???? That isn't even in the domain of the function. Do you perhaps mean $x_0=1$? I would not do this problem by calculating derivatives. Do you have any other techniques to consider? Nov 17 '19 at 17:33
• @TedShifrin Oh shoot, I made a typo. $x_0=1$.. Sorry, fixed it. And I'm not sure I know about any other ways. Could you help?
– user714814
Nov 17 '19 at 17:39
• Here's an easier one. Do you know how to give me the T.P. of degree $4$ centered at $x_0=0$ of $e^{x^2}$ without computing derivatives? (I assume you know the T.P. of $e^x$.) Nov 17 '19 at 17:41
• @TedShifrin Not sure how without calculating derivatives.
– user714814
Nov 17 '19 at 17:47
• You substitute $x^2$ for $x$ in the T.P. of degree $2$ of $e^x$. Can you figure out (prove?) why this works? Nov 17 '19 at 17:49

To make life a bit easier, what I would do is $$y=x^{x^x}\implies z=\log(y)=x\log(x^x)$$for which $$\frac {dz}{dx}=x^x \left(\frac{1}{x}+\log ^2(x)+\log (x)\right)\to 1$$ $$\frac {d^2z}{dx^2}= x^{x-2} (2 x+x \log (x) (x+x \log (x) (\log (x)+2)+3)-1)\to 1$$ For sure, this is tedious but doable.
We should arrive to $$z=(x-1)+\frac{1}{2} (x-1)^2+\frac{5}{6} (x-1)^3+\frac{1}{12} (x-1)^4+O\left((x-1)^5\right)$$ Now
$$y=e^{z}=1+(x-1)+(x-1)^2+\frac{3}{2} (x-1)^3+\frac{4}{3} (x-1)^4+O\left((x-1)^5\right)$$