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):

1. for $x<1,$ $y$ becomes 0
2. for $x=1,$ $y$ becomes 1
3. 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:

1. $x<1$ to be 0.
2. $x=1$ to be 1.
3. 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?

• Could you use, mathjax? :)
– Mann
Jun 8 '15 at 16:57
• Your correct observations A, B and C (given the particular way you interpret the infinite tower of exponents) tell you what the graph of $y$ as a function of $x$ looks like. It's the $x$-axis up to $1$, then it's $1$ at $1$, then it's undefined. So the derivative is $0$ up to $1$ and undefined after that. You don't need any derivative rules (or logarithm rules) for this problem. For the other interpretation you need to work more. Jun 8 '15 at 17:03
• Your observations A and C for the infinite power tower are not right. For example for $x=\sqrt2$ you ought to get $y=2$, not infinity. Quick test: with three layers, do you get $\sqrt 2^{\sqrt2^{\sqrt2}}=2$ or $\sqrt 2^{\sqrt2^{\sqrt2}}\approx 1.76$? The latter is right, because $x^{x^x}$ means $x^{(x^x)}$, not $(x^x)^x$. Jun 8 '15 at 17:03
• To compute the derivative, it is probably less confusing an error-prone to observe that $x^y=y$ implies $x=y^{1/y}$, and so the function you're looking at is the inverse of something that can be written down explicitly. The rule for the derivative of an inverse function will then give you the answer relatively painlessly. Jun 8 '15 at 17:05
• Your example of $x=2$ is too big. As Jack D'Aurizio says, you should only be considering $x \in \left[e^{-e},e^{\frac{1}{e}}\right] \approx [0.06598804, 1.44466786]$ Jan 6 '16 at 1:12

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.
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 $0$).