Finding second derivative of $x^{x^x}$ If $f(x) = x^{x^x} $, then find $f''(1)$.
My attempt:Now,
$\begin{align}y =x^{x^x} \\&\implies ln(y) = x^x \cdot ln(x)
 \\&\implies ln(ln(y))= x\cdot ln(x)+ln(ln(x))\\&\implies\frac{y'}{y\cdot ln(y)}
=ln(x) +1+\frac{1}{x\cdot ln(x)}\\&\implies y'=(x^{x^x})\cdot (x^xln(x))\cdot [ln(x) +1+\frac{1}{xln(x)}]
\end{align}$
I find it really tedious to continue in this way to reach desired answer.I have already checked finding second derivative of $x^x$. So, I wonder if I have to continue in the similar manner to get to answer or there is any better method to calculate this?

Proceeding along your approach
Let $y = x^{x^x}$. We have $y(1) = 1$.
We have
$$y' = y x^x (\ln^2 x + \ln x + x^{-1}).$$
We have $y'(1) = 1$.
We have
\begin{align*}
 y'' &= y' x^x (\ln^2 x + \ln x + x^{-1}) + y \cdot x^x(\ln x + 1) \cdot (\ln^2 x + \ln x + x^{-1})\\
 &\qquad + y x^x (2x^{-1}\ln x + x^{-1} - x^{-2} ).
\end{align*}
Letting $x = 1$, using $y(1) = 1$ and $y'(1)=1$, we have
$y''(1) = 2$, i.e. $f''(1) = 2$.
