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!

\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.
