Differentiate $\displaystyle y = a^{x ^{a^{x^\cdots}}}$. Could anyone help with this question:

Edit: Reducing it to one of these options would be great!
 A: 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.
A: \begin{align*}
y&=a^{x^{a^{x\cdots}}}\\
\implies y&=a^{x^y}\\
\implies \ln y&=x^y\ln a\hspace{25pt}\cdots\text{(i)}\\
\implies \dfrac{1}{y}\dfrac{dy}{dx}&=\ln a\cdot\dfrac{d}{dx}\left(x^y\right)\\
 &=\ln a\cdot\dfrac{d}{dx}\left(e^{y\ln x}\right)\\
&=\ln a\cdot e^{y\ln x}\cdot\dfrac{d}{dx}(y\ln x)\\
&=\ln a\cdot x^y\cdot\left(\dfrac{y}{x}+\ln x\cdot\dfrac{dy}{dx}\right)\\
\implies \dfrac{dy}{dx}&=y\cdot x^y\ln a\cdot\left(\dfrac{y}{x}+\ln x\cdot\dfrac{dy}{dx}\right)\\
\implies \dfrac{dy}{dx}\left(1-y\cdot x^y\ln a\cdot\ln x\right)&=\dfrac{y^2}{x}\cdot x^y\ln a\\
\implies \dfrac{dy}{dx}\left(1-y\cdot\ln y\cdot\ln x\right)&=\dfrac{y^2}{x}\cdot\ln y\hspace{25pt}\text{ as from (i) }[x^y\ln a=\ln y]\\
\implies \dfrac{dy}{dx}&=\dfrac{y^2\ln y}{x\left(1-y\cdot\ln y\cdot\ln x\right)}
\end{align*}
So, this doesn't matches any answer.
A: Building blocks:
$$
 D\left( a^{t(x)}, x\right) = a^x \ln (a) t'(x), \qquad
 D\left( x^x, x\right) = x^x (1 + \ln(x))
$$
First chains
$$
\begin{align}
  D\left( a^{x}, x\right) &= a^x \ln (a) \\
  D\left( a^{x^{x}}, x\right) &= x^x a^{x^x} \ln (a) (\ln (x)+1) \\
 D\left( a^{x^{x^{x}}}, x\right)&= a^{x^{x^x}} x^{x^x}  \ln (a) \left(x^{x-1}+x^x \ln (x) (\ln (x)+1)\right) \\
  D\left( \left( \left( \left( a^x \right)^x \right)^x \right)^x, x \right)
&= \left(\left(\left(a^x\right)^x\right)^x\right)^x \left(x \left(x \left(\ln \left(a^x\right)+x \ln (a)\right)+\ln \left(\left(a^x\right)^x\right)\right)+\ln \left(\left(\left(a^x\right)^x\right)^x\right)\right)
\end{align}
$$
