Find the derivative of $f(x)=x^{x^{\dots}{^{x}}}$. 
Find the derivative of $f(x)$:
$$f(x)=x^{x^{\dots}{^{x}}}$$

Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:
\begin{align}
n = 2 \Longrightarrow f(x) &= x^x \\
f'(x) &= \frac{d}{dx}\left(e^{x\ln \left(x\right)}\right) \\
&= e^{x\ln \left(x\right)}\frac{d}{dx}\left(x\ln \left(x\right)\right) \\
&= e^{x\ln \left(x\right)}\left(\ln \left(x\right)+1\right) \\
&= x^x\left(\ln \left(x\right)+1\right)
\end{align}
\begin{align}
n = 3 \Longrightarrow f(x) &= x^{x^{x}} \\
f'(x) &= \frac{d}{dx}\left(e^{x^x\ln \left(x\right)}\right) \\
&= e^{x^x\ln \left(x\right)}\frac{d}{dx}\left(x^x\ln \left(x\right)\right) \\
&= e^{x^x\ln \left(x\right)}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right) \\
&= x^{x^x}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)
\end{align}
\begin{align}
n = 5 \Longrightarrow f(x) &= x^{x^{x^{x^{x}}}} \\
f'(x) &= ... \\
&= x^{x^{x^{x^x}}}\left(x^{x^{x^x}}\ln \left(x\right)\left(x^{x^x}\ln \left(x\right)\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)+x^{x^x-1}\right)+x^{x^{x^x}-1}\right)
\end{align}
However, I am not sure if I see a pattern here that can help solve the question.
 A: All the other answers are for the case where there are infinite x's. However, there are finite ($n$) x's in this problem!
Let's first denote the function as $f_n(x)=x^{x^{\cdots^x}}$ with $n$ x's in the exponent. For example, $f_1(x)=x^x, f_0(x)=x.$
Take the logarithm of both sides, we get $\ln f_n(x)=f_{n-1}(x)\ln x$. Taking the derivative gives $$\frac{f'_n(x)}{f_n(x)}=f'_{n-1}(x)\ln x+\frac{f_{n-1}(x)}{x}$$
$$f'_n(x)=f_n(x) f'_{n-1}(x)\ln x+\frac{f_n(x) f_{n-1}(x)}{x}$$
That's just a basic recurrence relation. We know that $$a_n=Ca_{n-1}+D\implies a_n=c C^{n-1} + \frac{D(C^n-1)}{C-1}$$ where $c$ is an arbitrary constant.
Similarly, we may get $$f'_n(x)=c(f_n(x)\ln x)^{n-1}+\frac{f_n(x)f_{n-1}(x)}{x(f_n(x)\ln x-1)} ((f_n(x)\ln x)^n-1)$$
To get the value of the parameter $c$, let $n=1$, then $f'_1(x)=c+x^x$. Hence, $c=x^x \ln x$. Substituting in, we get the final answer (in a rather cumbersome form) $$f'_n(x)=x^x \ln x(f_n(x)\ln x)^{n-1}+\frac{f_n(x)f_{n-1}(x)}{x(f_n(x)\ln x-1)} ((f_n(x)\ln x)^n-1)$$
A: Let $f_1(x)=x$ and $f_n(x)=x^{f_{n-1}(x)}=x^{x^{\ldots^{x}}}$ ($n$ $x's$ on the exponent).
Then $\ln(f_n(x))=f_{n-1}(x)\cdot\ln(x)$ so: $$\frac{f'_n(x)}{f_{n}(x)}=f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\implies f'_{n}(x)=f_{n}(x)\cdot\left(f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\right).$$
A: $$
f(x) = x^{f(x)}\\
\ln(f(x))=f(x) \ln(x)\\
\frac{1}{f(x)}f'(x) = \frac{f(x)}{x} + \ln(x)f'(x),\\
f'(x)=\left(\frac{1}{f(x)}-\ln(x)\right)^{-1}\frac{f(x)}{x},\\
f'(x)=\left(\frac{f(x)}{1-\ln(x)f(x)}\right)\frac{f(x)}{x}.
$$
A: $$x^{x^{x^\cdots}}= y \implies x^y = y \implies1 = yx^{-y} = ye^{-y\log (x)}$$
which means
$$y = \frac{-W(-\log (x))}{\log (x)}$$
where $W$ is the principal branch of the Lambert-W function. Now, the problem of the successive derivatives is "much" simpler (have a look here for the derivatives of $W(t)$).
