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Hope you’re still fine. Let’s just call that “derivatives gone wild.” 🙂

ReplyBy Math Vault | Calculus

In a first course of calculus, we have very specific rules for dealing with functions in the form of a sum, a difference, a product and a quotient. And now, we’re about to throw a curve ball at you:

## What is the derivative of the function $f(x)=x^{x^x}$?

Intrigued? Well, it’s time to learn more! For we can assure you there’s a whole world that haven’t been much explored before!

While **polynomial** (e.g., $x^3$) and **exponential** functions (e.g., $\pi^x$) both have relatively easy derivatives, such is *not* the case with functions such as $x^x$, where the variable occurs both in the **base** and in the **exponent**. However, there is a standard trick in these cases, which is to apply **logarithm** and **exponentiation** (of the *same* base, and in that order) to the original function, subsequently turning it into an *equivalent function* which then can be differentiated later.

In the case of $f(x)=x^x$, applying logarithm and exponentiation (with base $e$) yields the following equation:

$$x^x=e^{\ln{(x^x)}}$$

This is a valid step, and is warranted by the fact that natural logarithm and natural exponentiation are **inverse operations**. Note, however, that the result is *only* applicable insofar as $x^x>0$ (e.g., if we restrict the domain of $x^x$ to *only* the positive numbers).

And since $\ln{(x^x)} = x\ln{x}$ (by the **power rule** of logarithm), we also have that:

$$e^{\ln{(x^x)}}=e^{x\ln{x}}$$

In other words, *assuming that* $x>0$, the functions $x^x$ and $e^{x\ln{x}}$ can be used *interchangeably*, so that differentiating the former is akin to differentiating the latter, which in turn can be carried out through a combination of **chain rule** and **product rule**:

$$(e^{x\ln{x}})’=e^{x\ln{x}}(x\ln{x})’=e^{x\ln{x}}[(x)’\ln{x}+x(\ln{x})’]=e^{x\ln{x}}[\ln{x}+1]$$

And since $x^x$ and $e^{x\ln{x}}$ are **equivalent functions** (again, assuming that $x>0$), replacing each occurrence of $e^{x\ln{x}}$ back into $x^x$, we get:

$$(x^x)’=x^x(\ln{x}+1)$$

Moral of the story? Much like the same way differentiating $a^x$ makes an additional **multiplier** pop out (i.e., $\ln{a}$), differentiating $x^x$ also makes an additional term pop out (in this case, $\ln{x}+1$).

**Summary**

The derivative of $x^x$ is $x^x(\ln{x}+1)$, for any $x>0$.

Note that the derivative of $x^x$ can also computed (albeit in a roundabout fashion) using the so-called **logarithmic differentiation**, as Sal illustrated here in this video:

Alternatively, the same derivative can also be obtained using the **Exponent Rule**, a powerful differentiation rule for handling generalized powers.

Before going on any further though, it’s imperative for us to determine what the expression $x^{x^x}$ actually refers to. For example, when we say $2^{1^2}$, are we talking about ${(2^1)}^2$? Or maybe $2^{(1^2)}$? To be sure, when the parentheses are on the *left*, we get ${(2^1)}^2$, which is equal to $2^2=4$, but when the parentheses are on the *right*, we get $2^{(1^2)}$, which is equal to $2^1=2$.

Oops. Not the same at all! 😳

In other words, whereas addition and multiplication are both **associative** (i.e., $(a+b)+c=a+(b+c), (ab)c=a(bc)$), the same is *not true* for **exponentiation**, which could result in a potential misuse of notation if we allow ourselves to be a bit sloppy:

$${(a^b)}^c \ne a^{(b^c)}$$

An immediate corollary from this would be that the parentheses constitute a *necessar*y component in preserving the well-defineness of expressions involving *several* exponentiation operators. However, because humans are generally terrible in *parsing* symbols (and prefer to avoid them altogether if possible), there is a **convention** out there which stipulates that when no parenthesis is written, we may assume that they are on the* right-hand side*. That is:

$a^{b^c}\stackrel{def}{=}a^{(b^c)}$ ; $a^{b^{c^d}}\stackrel{def}{=}a^{({b^{({c^d})}})}$ ; $\ldots$

**Translation**

When the parentheses are on the left (e.g., ${(a^b)}^c$), they *cannot* be dropped.

There is yet another source of ambiguity that needs to be addressed. That is, what happens when $x$ is not positive (e.g., $x=-\frac{1}{2}$)? You see, the problem is that $(-\frac{1}{2})^{-\frac{1}{2}}=\frac{1}{({-\frac{1}{2}})^{\frac{1}{2}}}=\frac{1}{\sqrt{-\frac{1}{2}}}$, so it cannot be defined without throwing in some **complex numbers**. Let alone $(-\frac{1}{2})^{{-\frac{1}{2}}^{-\frac{1}{2}}}$…

To prevent potential complications with **negative bases**, it is therefore customary to *restrict* the domain of $x^{x^x}$ to the set of **positive** numbers.

In fact, this will be our de-facto domain for any function of the form $x^{x^{x^x\dots}}$.

**Recap**

For functions such as $x^{x^{x^x\dots}}$, it’s generally assumed that $x>0$ and that parentheses are placed on the *right* — unless otherwise specified.

Now that we have cleared out some potential roadblocks, we can proceed to differentiate ${(x^x)}^x$ — using the *same* technique we employed in differentiating $x^x$.

First, turning ${(x^x)}^x$ into an **equivalent exponential function**, we get:

$${{(x^x)}^x}=e^{\ln{[{(x^x)}^x}]}=e^{x\ln{(x^x)}}$$

Differentiating the equivalent exponential function using a combination of **chain rule** and **product rule** yields:

\begin{align*} (e^{x\ln{(x^x)}})’ & = e^{x\ln{(x^x)}}(x\ln{(x^x)})’ \\ & =e^{x\ln{(x^x)}}[\ln(x^x)+x\frac{1}{x^x}(x^x)’]\end{align*}

Since we found earlier that $(x^x)’=x^x(\ln{x}+1)$, we have:

\begin{align*} (e^{x\ln{(x^x)}})’ & = e^{x\ln{(x^x)}}[\ln(x^x)+x\frac{1}{x^x}(x^x)(\ln{x}+1)] \\ & = e^{x\ln{(x^x)}}[\ln(x^x)+x(\ln{x}+1)] \end{align*}

Finally, replacing each occurrence of $e^{ x\ln{(x^x)} }$ back into ${{(x^x)}^x}$, we get:

\begin{align*} ({{(x^x)}^x})’ & = {{(x^x)}^x}[\ln(x^x)+x(\ln{x}+1)] \\ & = {{(x^x)}^x}[x\ln{x}+x\ln{x}+x)] \end{align*}

Finally, simplifying further yields…

$$({{(x^x)}^x})’={{(x^x)}^x}x(2\ln{x}+1)$$

Neat. Moving on!

Same drill. First:

$$x^{(x^x)}=e^{\ln{[x^{(x^x)}]}} = e^{(x^x)\ln{x}}$$

Second, differentiate:

\begin{align*}[e^{(x^x)\ln{x}}]’ & = [e^{(x^x)\ln{x}}]{[(x^x)\ln{x}]}’ \\ & =[e^{(x^x)\ln{x}}] [(x^x)’\ln{x}+(x^x)\frac{1}{x}] \end{align*}

Third, expand $(x^x)’$ and simpify:

\begin{align*} [e^{(x^x)\ln{x}}]’ & = [e^{(x^x)\ln{x}}] [(x^x)(\ln{x}+1)\ln{x}+\frac{x^x}{x}] \\ & =[e^{(x^x)\ln{x}}] [(x^x)(\ln{x}+1)\ln{x}+x^{x-1}] \end{align*}

Finally, replace each occurrence of $e^{(x^x)\ln{x}}$ back into $x^{(x^x)}$:

$$(x^{(x^x)})’ = (x^{(x^x)})[(x^x)(\ln{x}+1)\ln{x}+x^{x-1}]$$

Got it!

Note that alternatively, the derivative can also be computed again using** logarithmic differentiation**, as Sal illustrated in this video:

**Summary**

Given that $x>0$, the derivative of ${{(x^x)}^x}$ is ${{(x^x)}^x}x(2\ln{x}+1).$ The derivative of $x^{(x^x)}$ is a bit longer though.

So this leads us to the natural question: what about $x^{x^{x^x}}$? Or maybe ${({(x^x)}^x)}^x$? But first of all, we can’t be writing $x$ forever can’t we? And even if you can, it would be hard for us to *type* them out with all the coding…

With that in mind, let’s introduce a new notation — a new shorthand specifically geared towards **iterated exponentiation**:

$$\displaystyle ^nx \stackrel{def}{=} \underbrace{x^{x^{x^x\ldots}}}_{n \text{ times}} \quad (x>0, n \in \mathbb{N}) $$

For example. $^1x = x$, $^2\pi = \pi^\pi$, $^33.5=3.5^{3.5^{3.5}}=3.5^{(3.5^{3.5})}$, and $^4x=x^{x^{x^x}}=x^{(x^{(x^x)})}$ (remember that by convention, the parentheses are on the *right* by default).

The advantage of this notation, is that it comes in handy when we wish to talk about functions of the form $f(x)= {^n}x= \underbrace{x^{x^{x^x\ldots}}}_{n \text{ times}}$, which we shall henceforth refer to as **hyperexponential functions**. That is, they go beyond the exponential functions, and are constructed through an algebraic operation called **tetration **(i.e., the fourth **hyperoperation** — after **addition**, **multiplication** and **exponentiation**):

\begin{align*} x + n & \stackrel{def}{=} x + \underbrace{1 +1+\ldots +1}_{n \text{ times}}\\ x \times n & \stackrel{def}{=} \underbrace{x + x + \ldots + x}_{n \text{ times}} \\ x^n & \stackrel{def}{=} \underbrace{x \times x \times \ldots \times x}_{n \text{ times}} \end{align*}

In fact, we’ve already learnt that:

$({^1}x)’ = x’ = 1$

$({^2}x)’ = {(x^x)}’ = {^2}x(\ln{x}+1)$

$({^3}x)’ = {(x^{x^x})}’ = {^3}x[{^2}x(\ln{x}+1)\ln{x}+x^{x-1}]$

So the question becomes: is there a *neat*, **general formula** for $({^n}x)’$? Well, there is definitely one, but whether it’s neat is up for debate…

To see how, note that ${^n}x=e^{\ln{({^n}x})}=e^{({^{n-1}}x) \ln{x}}$. This means that *after differentiation*, we can express the derivative of ${^n}x$ is terms of the derivative of ${^{n-1}x}$. So a **recursive formula** for the derivative of ${^n}x$ is in order:

$$({^n}x)’={^n}x [({^{n-1}}x)’ \ln{x} + ({^{n-1}}x) \frac{1}{x}]$$

In particular, when $n=4$, we get:

$$({^4}x)’={^4}x [({^{3}}x)’ \ln{x} + ({^{3}}x) \frac{1}{x}]$$

Using the fact that $({^3}x)’ = {^3}x[{^2}x(\ln{x}+1)\ln{x}+x^{x-1}]$, we obtain a **close-form expression** for the derivative of ${^4}x$:

$$({^4}x)’={^4}x \{{^3}x[{^2}x(\ln{x}+1)\ln{x}+x^{x-1}] \ln{x} + ({^{3}}x) \frac{1}{x}\} $$

And this is also how we can figure out $({^5}x)’$, $({^6}x)’$ — or $({^{666}}x)’$, for that matter.

**Recap**

The derivative of ${^n}x$ can be determined *recursively* for all $n \in \mathbb{N}$, but it gets longer every time.

And of course, the show doesn’t stop here either. In fact, by iterating a series of tetrations, the fifth hyperoperation — **pentation** — can be developed as well, leading to even more fancy mathematical goodness!

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Math Vault January 3, 2016

Hope you’re still fine. Let’s just call that “derivatives gone wild.” 🙂

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