derivative of x^x^x... to infinity? I am a 12th grade student, and I am afraid that in realistic terms this question might not even make sense because of the infinities that have to be dealt with. However, in my attempt to calculate it's derivative, I did the following:
$$y=x^{x^{x^{.^{.^.}}}}$$
$$\ln(y)=\ln(x)x^{x^{x^{.^{.^.}}}}$$
$$\ln(y)=y\ln(x)$$
after taking derivative with respect to x on both sides, I obtained the following:
$$\frac{\mathrm dy}{\mathrm dx}=\frac{y^2}{x(1-\ln(y))}$$
I am fairly certain that the calculations up to this point are valid. However, to further continue, I analyzed the nature of y in different domains and obtained the values, by observation(that is, by observing how $x$, $x^x$, $x^{x^x}$, $x^{x^{x^x}}$, and so on would behave to draw a conclusion):


*

*for $x<1,$ $y$ becomes 0

*for $x=1,$ $y$ becomes 1

*for $x>1,$ $y$ becomes infinite


These 3 points is where the first problem lies. Are these true? If so, how do we reach to the conclusion?
Secondly, considering this to be true,
I get the derivative at:


*

*$x<1$ to be 0.

*$x=1$ to be 1.

*for $x>1$, I took $x=2.$ then the derivative $\frac{\mathrm dy}{\mathrm dx} = y^2/[2(1-ln(y))]$ (replacing $x$ by $2$). Now, I applied L hospital's rule to get the value of the expression to be negative infinity.


This is the second problem. I have used L Hospital's rule, but limits were not concerned. Is this method valid? If not, how would we calculate it?
 A: By following this link you will see that you function is well-defined over the interval $I=\left[e^{-e},e^{\frac{1}{e}}\right]$ and the values in the endpoints are just $\frac{1}{e}$ and $e$. Over such interval, the functional equation:
$$ f(x) = x^{f(x)} \tag{1}$$
leads to:
$$ f'(x) = x^{f(x)}\left(f'(x)\log x+\frac{f(x)}{x}\right)\tag{2}$$
hence:
$$ f'(x) = \frac{f(x)^2}{x\left(1-f(x)\log x\right)}\tag{3}$$
as you stated. We may also notice that:
$$ f(x)=\frac{W(-\log x)}{-\log x}\tag{4}$$
where $W$ is the Lambert W-function.
A: Not really an anser, rather an outline of what I would try to study this.
The first problem, at least to me, is the definition of the function (what you called "dealing with the infinite". The way I would attempt to define it is saying that the function $f$ is implicitly defined by the equation
$$\begin{equation}
f(x) = e^{f(x)\ln x}, \quad x > 0.
\end{equation}$$
Now, the second problem is arguing why this function exists and is continuous, let alone differentiable. Note that if it exists, you automatically have a nice property: $f> 0$ on its domain of definition.
Assuming this is done, you would need to argue that $f$ is differentiable, before actually deriving it. As a side remark, looking at $x=e^{1/e}$ in the equation defining $f$, you get that $f(e^{1/e})$ must be equal to $e$, so that $f$ cannot be differentiable at $e^{1/e}$ (otherwise, the expression of the derivative you get will be infinite -- division by $0$).
