TITLE: Differentiate $\displaystyle y = a^{x ^{a^{x^\cdots}}}$.
QUESTION [7 upvotes]: Could anyone help with this question:

Edit: Reducing it to one of these options would be great!

REPLY [6 votes]: I will assume that everything is well defined and real, not paying attention to whether logarithms are taken of positive numbers and such.
We have $y=a^{x^y}$ (not $y=(a^x)^y$ which would mean $y=a^{{xa}^{{xa}^{\cdots}}}$).
Let me denote the derivative operator by $D$.
First we need a little lemma:
If $f$ and $g$ are functions of $x$, then
$$
D(f^g)
=
De^{g\log(f)}
=
e^{g\log(f)}D(g\log(f))
=
f^g(g'\log(f)+gf'/f).
$$
(There is an easy way to memorize this. If $f$ is constant, the derivative is $f^g\log(f)g'$. If $g$ is constant, the derivative is $gf^{g-1}$. The full derivative is the sum of these two. This argument can be formalized, but it'd be a sidetrack here.)
Applying this to $f(x)=x$ and $g(x)=y(x)$ gives
$$
D(x^y)
=
x^y(y'\log(x)+y/x).
$$
This will be useful soon.
Taking the derivative gives
\begin{align}
y'
&=
D(a^{x^y})
\\&=
a^{x^y}\log(a)D(x^y)
\\&=
y\log(a)D(x^y)
%\\&=
%y\log(a)[yx^{y-1}+\log(x)x^yy']
\\&=
y\log(a)x^y[y/x+\log(x)y']
\\&=
y\log(a^{x^y})[y/x+\log(x)y']
\\&=
y\log(y)[y/x+\log(x)y']
.
\end{align}
This gives
$$
[1-\log(x)y\log(y)]y'
=
y^2\log(y)/x,
$$
from which we can solve
$$
y'=\frac{y^2\log(y)}{x[1-\log(x)y\log(y)]}.
$$
This does not seem to match any of the options given.
Here is an alternative form, using $x^y=\log(y)/\log(a)$:
$$
y'=\frac{y^2\log(y)}{x\left[1-\log\left(\frac{\log(y)}{\log(a)}\right)\log(y)\right]}.
$$
Perhaps I miss a way to manipulate the formula, or perhaps the problem is mistaken; as others have pointed out, taking $y=(a^x)^y$ leads to option C.