Exponent:
$f^g = exp(log(f^g))$
$= exp(g \cdot log(f))$

Exponent Derivative:
$( f^g )' = ( exp(g \cdot log(f))'$
$= exp(g \cdot log(f)) \cdot (g \cdot log(f))'$
$= exp(g \cdot log(f)) (g' \cdot log(f) + g \cdot log'(f))$
$= f^g (g' \cdot log(f) + g f' \frac{1}{f})$

Exponent Derivative of Integer Power:
$g = n, g' = 0, f = x, f' = 1$
$(f^n)' = f^n \cdot \frac{n}{x}$
$= n f^{n-1}$

Exponent Derivative of $x^x$
$f = g = x, f' = g' = 1$
$(x^x)' = x^x (log(x) + x \frac{1}{x})$
$= x^x (log(x) + 1)$

---

Tetration:
${}^n x = \underbrace{{x^{x^{...{}^x}}}}_{\times n}$

Tetration Derivative:
$({}^n x)'$
$ = (\underbrace{{x^{x^{...{}^x}}}}_{\times n})'$
$ = \underbrace{(exp(x \cdot log( exp(x \cdot log( ... (exp( x \cdot log(x))) ... )) )))'}_{\times n}$

Looks too tedious for a chain rule. Let's try a recursive approach.

$({}^0 x) = 1$
$({}^0 x)' = 0$

$({}^1 x) = x$
$({}^1 x)' = x' = 1$

$({}^n x)' = (x^{({}^{(n-1)} x)})'$
$ = (exp ( log(x) \cdot {}^{(n-1)} x ))'$
$ = exp ( log(x) \cdot {}^{(n-1)} x ) \cdot ( log(x) \cdot {}^{(n-1)} x )'$
$ = log(x) \cdot {}^{n} x \cdot ( ({}^{(n-1)} x)' + \frac{1}{x} \cdot {}^{(n-1)} x )$
$ = log(x) \cdot {}^{n} x \cdot ({}^{(n-1)} x)' + \frac{1}{x} \cdot {}^{n} x \cdot {}^{(n-1)} x $
... hmm, at this point I'm going to substitute again...
$ = log(x) \cdot {}^{n} x \cdot ( log(x) \cdot {}^{(n-1)} x \cdot ({}^{(n-2)} x)' + \frac{1}{x} \cdot {}^{(n-1)} x \cdot {}^{(n-2)} x ) + \frac{1}{x} \cdot {}^{n} x \cdot {}^{(n-1)} x$
$ = (log(x))^2 \cdot {}^{n} x \cdot {}^{(n-1)} x \cdot ({}^{(n-2)} x)' + \frac{1}{x} \cdot {}^{n} x \cdot {}^{(n-1)} x \cdot {}^{(n-2)} x \cdot log(x) + \frac{1}{x} \cdot {}^{n} x \cdot {}^{(n-1)} x$
... substitute again ...
$ = (log(x))^2 \cdot {}^{n} x \cdot {}^{(n-1)} x \cdot ( log(x) \cdot {}^{(n-2)} x \cdot ({}^{(n-3)} x)' + \frac{1}{x} \cdot {}^{(n-2)} x \cdot {}^{(n-3)} x ) + \frac{1}{x} \cdot {}^{n} x \cdot {}^{(n-1)} x \cdot {}^{(n-2)} x \cdot log(x) + \frac{1}{x} \cdot {}^{n} x \cdot {}^{(n-1)} x$
$ = (log(x))^3 \cdot {}^{n} x \cdot {}^{(n-1)} x \cdot {}^{(n-2)} x \cdot ({}^{(n-3)} x)' + \frac{1}{x} \cdot (log(x))^2 \cdot {}^{n} x \cdot {}^{(n-1)} x \cdot {}^{(n-2)} x \cdot {}^{(n-3)} x + \frac{1}{x} \cdot {}^{n} x \cdot {}^{(n-1)} x \cdot {}^{(n-2)} x \cdot log(x) + \frac{1}{x} \cdot {}^{n} x \cdot {}^{(n-1)} x$

Hypothesis:
$({}^n x)' = \frac{1}{x} \cdot \Sigma_{j=0}^{(n-1)} \left( (log(x))^{j} \cdot \Pi_{k=n-j}^n \left( {}^k x \right) \right) $
...not quite right.

---

$({}^1 x)' = 1$
$({}^2 x)' = {}^2 x \cdot (log(x) + 1) $
$({}^3 x)' = (x^{x^x})' = {}^3 x \cdot ( {}^2 x \cdot (log (x) + x \cdot \frac{1}{x}) \cdot log(x) + {}^2 x \cdot \frac{1}{x}) = {}^3 x \cdot {}^2 x \cdot ( log(x)^2 + log(x) + \frac{1}{x})$
$({}^4 x)' = (x^{x^{x^x}})' = {}^4 x \cdot {}^3 x \cdot {}^2 x \cdot ( log(x)^3 + log(x)^2 + log(x) \cdot \frac{1}{x}) + {}^4 x \cdot {}^3 x \cdot \frac{1}{x} $