Differentiation of exponential function? How to solve derivative $\lim_{n\to\infty}e^{{}^n(x)}$ with respective of $x$ ? Here, ${}^n(x)$ is a tetration function
$$
{}^n(x)=
\begin{cases}
 x^{[{}^{n-1}(x)]}  & \mbox{ if  } {\;n>1}\\
 x                   & \mbox{ if  } {\;n=1}\\
\end{cases}
$$
Anyone give me any idea how can resolve this problem ?
Thanks
 A: If you have meant $$\displaystyle e^{(x^{x^{\cdots\infty}})}$$
I will write $$\displaystyle  x^{x^{\cdots\infty}}=y\implies x^y=y\ \ \ \ (1)$$
Now, the problem becomes $$\frac{d(e^y)}{dx}$$ $$\text{ which is   }=e^y\cdot\frac{dy}{dx}$$
Can you find $\displaystyle\frac{dy}{dx}$ from $(1)$
A: The derivative $e^{{}^n(x)}$ will be given by some recursive formula. First all note that, for $n\geq 2$ we have that
\begin{align}
 \log\Big({}^n(x)\Big)=&  x\cdot \log\Big({}^{n-1}(x)\Big)\\
 = &  xx\cdot \log\Big({}^{n-2}(x)\Big)\\
= &  xxx\cdot \log\Big({}^{n-3}(x)\Big)\\
\vdots\;& \hspace{2cm}\vdots \\
= &  x^k\cdot \log\Big({}^{n-k}(x)\Big), \quad k=1,\ldots n-1\\
\vdots\;& \hspace{2cm}\vdots \\
= &  x^{n-1}\cdot \log\Big({}^{1}(x)\Big), \\
= &  x^{n-1}\cdot \log(x), \\
\end{align}
implies 
\begin{align}
D_x\Big({}^{n}(x)\Big)=&D_x\big(e^{\log\big({}^{n}(x)\big)}\big)\\
=& e^{\log\big({}^{n}(x)\big)}D_x\big( x^{n-1}\cdot \log(x) \big)\\
=& {}^{n}(x)\cdot \bigg( (n-1)\cdot x^{n-2}\cdot\log(x)+x^{n-1}\cdot \frac{1}{x} \bigg)\\
=& {}^{n}(x)\cdot \bigg( (n-1)\cdot x^{n-2}\cdot\log(x)+x^{n-2} \bigg)\\
=& {}^{n}(x)\cdot x^{n-2}\cdot\bigg( (n-1)\cdot \log(x)+1 \bigg)\\
\end{align}
Finally,
$$
D_x\Big[ e^{{}^n(x)}\Big]=e^{{}^n(x)}\cdot {}^{n}(x)\cdot x^{n-2}\cdot\bigg( (n-1)\cdot \log(x)+1 \bigg)
$$
But what we want is derivative of 
$$
D_x\Big[ \lim_{n\to\infty}e^{{}^n(x)}\Big]
%=e^{{}^n(x)}\cdot {}^{n}(x)\cdot x^{n-2}\cdot\bigg( (n-1)\cdot \log(x)+1 \bigg)
$$
So we need to ensure the convergence of series of functions. The ideal would be uniform rather than just pointwise convergence convergence. In uniform convergence we have 
$$
D_x\Big[ \lim_{n\to\infty}e^{{}^n(x)}\Big]
=\lim_{n\to \infty}
\left[
e^{{}^n(x)}\cdot {}^{n}(x)\cdot x^{n-2}\cdot\bigg( (n-1)\cdot \log(x)+1 \bigg)
\right]
$$
It's easy to see that sequence $\{e^{{}^n(x)} \}_n$ is monotone in itervals $(0,1),(1,\infty)$. If  $\{e^{{}^n(x)} \}_n$  converge monotonically in any this intervals we use the  Dini's theorem

THEOREM(Dini) Let $f_n\to f$ pointwise and monotonically over $[a,b]$, with each $f_n$ continuous, and $f$ continuous. Then $f_n\to f$ uniformly.

to conclude
$$
D_x\Big[ \lim_{n\to\infty}e^{{}^n(x)}\Big]
=\lim_{n\to \infty}
\left[
e^{{}^n(x)}\cdot {}^{n}(x)\cdot x^{n-2}\cdot\bigg( (n-1)\cdot \log(x)+1 \bigg)
\right]
$$
Analyse the pontual convergence of $\lim_{n\to\infty}e^{{}^n(x)}$. 


*

*For $x> 1$  the limit is infinite.

*For $0 <x <1$ the limit is zero $\lim_{n\to\infty}e^{{}^n(x)}=0$ and converge monotonically.


What can you conclude?
