Derivative of xxxx^{x^x}????

Differentiating xxxx^{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 xxx^x? Let's give it a shot!

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

ddx(xx)=ddx(exlnx)=exlnxddx(xlnx)\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 xlnxx\ln x as well but that is no big deal: ddx(xlnx)=lnx+1    exlnxddx(xlnx)=exlnx(lnx+1)=xx(lnx+1)=xxlnx+xx\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 ddx(xx)=xxlnx+xx\cfrac{d}{dx}(x^x) = x^x \ln x + x^x, we can move onto the main topic of this note: the derivative of xxxx^{x^x}!

We can start off the same as we did with the warmup: ddx(xxx)=ddx(exxlnx)=exxlnxddx(xxlnx)\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 xxx^x again. Good thing we did it earlier! ddx(xxlnx)=lnx(xxlnx+xx)+xxx=xxln2(x)+xxlnx+xx1=xx1(xln2(x)+xlnx+1)\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}

    exxlnxddx(xxlnx)=xxx(xx1(xln2(x)+xlnx+1))=xxx+x1(xln2(x)+xlnx+1)\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: ddx(xxx)=xxx+x1(xln2(x)+xlnx+1)\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 xxxxx^{x^{x^x}} is doable but will take time and a lot of xx's, and differentiating xxxxxx^{x^{x^{x^x}}} will take even more. I decided to do a quick generalisation in the end: ddx(nx)=nxddx(n1xlnx) where nx=xxxxxxno. of x’s=n\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 ef(x)e^{f(x)} being ef(x)f(x)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
1 month, 4 weeks ago

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