# Derivative of $x^{x^x}$????

Differentiating $x^{x^x}$? Is it worth doing? Yes because I can.

First of all, we should start off smaller with a warm up (it may help us later...). How about differentiating $x^x$? Let's give it a shot!

We can write $x^x$ as $e^{x\ln x}$. This makes our calculations a bit easier to do:

$\frac{d}{dx}\left(x^x\right) = \frac{d}{dx}\left(e^{x\ln x}\right) = e^{x\ln x} \cdot \frac{d}{dx}(x\ln x)$

Because of the chain rule, we need to differentiate $x\ln x$ as well but that is no big deal: \begin{aligned}\frac{d}{dx}(x\ln x) = \ln x + 1 \implies e^{x\ln x} \cdot \frac{d}{dx}(x\ln x) &= e^{x\ln x} (\ln x + 1) \\ &= x^x(\ln x + 1) \\ &= \green{\boxed{x^x \ln x + x^x}} \end{aligned}

So, now that we know that $\cfrac{d}{dx}(x^x) = x^x \ln x + x^x$, we can move onto the main topic of this note: the derivative of $x^{x^x}$!

We can start off the same as we did with the warmup: $\frac{d}{dx}\left(x^{x^x}\right) = \frac{d}{dx}\left(e^{x^x \ln x}\right) = e^{x^x \ln x} \cdot \frac{d}{dx}\left(x^x \ln x\right)$

Once again, because of the chain rule, we need to differentiate the exponent here too. It also looks like we need to differentiate $x^x$ again. Good thing we did it earlier! \begin{aligned} \frac{d}{dx}\left(x^x \ln x\right) &= \ln x (x^x \ln x + x^x) + \frac{x^x}{x} \\ &= x^x \ln^2(x) + x^x \ln x + x^{x-1} \\ &= x^{x-1}(x\ln^2(x) + x\ln x + 1) \end{aligned}

\begin{aligned} \implies e^{x^x \ln x} \cdot \frac{d}{dx}\left(x^x \ln x\right) &= x^{x^x}\left(x^{x-1}(x\ln^2(x) + x\ln x + 1)\right) \\ &= \green{\boxed{x^{x^x + x - 1}(x\ln^2(x) + x\ln x + 1)}} \end{aligned}

So there we have it: $\cfrac{d}{dx}\left(x^{x^x}\right) = x^{x^x + x - 1}(x\ln^2(x) + x\ln x + 1)$ But can we do better? Differentiating $x^{x^{x^x}}$ is doable but will take time and a lot of $x$'s, and differentiating $x^{x^{x^{x^x}}}$ will take even more. I decided to do a quick generalisation in the end: $\frac{d}{dx}({^{n}x}) = {^{n}x} \cdot \frac{d}{dx}\left({^{n-1}x}\ln x\right) \text{ where }{^{n}x} = \underbrace{x^{x^{x^{x^{x^{x^\dots}}}}}}_{\text{no. of }x \text{'s} = n}$

It is quite nice how it is similar (kind of) to the derivative of $e^{f(x)}$ being $e^{f(x)} \cdot f'(x)$ which shows that it is a special function itself.

I hope you enjoyed this note if you took the time to read it!

Note by James Watson
11 months, 1 week ago

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this explanation is too complicated and difficult for me

- 6 months, 3 weeks ago

ok

- 6 months, 3 weeks ago