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
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
$$y=e^{z}=1+(x-1)+(x-1)^2+\frac{3}{2} (x-1)^3+\frac{4}{3} (x-1)^4+O\left((x-1)^5\right)$$