Derivative of nx{^{n}x}?? (Continuation of my previous note)

In my last note, I went over the derivatives of xxx^x and xxxx^{x^x}. At the end, I went over a brief generalisation of differentiating nx{^{n}x}: ddx(nx)=nxddx(n1xlnx)\frac{d}{dx}({^{n}x}) = {^{n}x} \cdot \frac{d}{dx}\left({^{n-1}x}\ln x\right) However, I am not satisfied with this because this is not fully calculated (you still need to differentiate n1xlnx{^{n-1}x}\ln x and so on). I want to have a nice expression for this.

Before we start picking apart this problem, I want to define a new function (unless it isn't new) because it may or may not be useful later. Let g(x)=f(x)(f(x)(f(x)((f(x)+t)+t)+t))+tg(x) = f(x)(f(x)(f(x)(\dots(f(x)+t)+t)+t)\dots)+t where f(x)f(x) is another function and kk is some constant. There could be nn many brackets here. For the purposes of this note, I want to notate this as g(x)=Bk=0nf(x)β+tg(x) = \overset{\small n}{\underset{\small k=0}{\huge \Beta }}f(x) \beta + t Here, β\beta is where the nested brackets will go. Here is an example: Bk=02aβ+b=a(a(a+b)+b)+b\overset{\small 2}{\underset{\small k=0}{\huge \Beta }}a \beta + b = a(a(a+b)+b)+b Here is another: Bk=36kβt=6(5(4(3t)t)t)t\overset{\small 6}{\underset{\small k=3}{\huge \Beta }}k\beta-t = 6(5(4(3-t)-t)-t)-t

Hopefully you understand what this function does. Anyway, onto the main topic of this note!

Let us be reminded of our starter last note: the derivative of xxx^x. We found out that it was equal to xx(lnx+1)x^x(\ln x + 1) We also found out that the derivative of xxxx^{x^x} was xxx+x1(xln2(x)+xlnx+1)x^{x^x + x - 1}(x\ln^2(x) + x\ln x + 1) which can also be written as xxx+x1(xln(x)(ln(x)+1)+1)x^{x^x+x-1}\left(x\ln\left(x\right)\left(\ln\left(x\right)+1\right)+1\right) For the sake of this note, I will not be calculating the derivative of xxxxx^{x^{x^x}} by hand or any higher tetrations. If we look at the derivative of xxxxx^{x^{x^x}}, it turns out to be xxxx+xx1(xxln(x)(xln(x)(ln(x)+1)+1)+1)x^{x^{x^x}+x^x-1}\left(x^x\ln\left(x\right)\left(x\ln\left(x\right)\left(\ln\left(x\right)+1\right)+1\right)+1\right) Maybe you start to see a pattern here. Just to check that there is one, we can observe the derivative of xxxxxx^{x^{x^{x^x}}}: xxxxx+xxx1(xxxln(x)(xxln(x)(xln(x)(ln(x)+1)+1)+1)+1)x^{x^{x^{x^x}}+x^{x^x}-1}\left(x^{x^x}\ln\left(x\right)\left(x^x\ln\left(x\right)\left(x\ln\left(x\right)\left(\ln\left(x\right)+1\right)+1\right)+1\right)+1\right)

That's a lot of xx's! However, in this giant mess, there is a pattern that begins to appear.

Can we try expressing these derivatives as the function that I established earlier? Let us do xxxxx^{x^{x^x}}: xxxx+xx1(xxln(x)(xln(x)(ln(x)+1)+1)+1)=xxxx+xx1Bk=02kxln(x)β+1x^{x^{x^x}+x^x-1}\left(x^x\ln\left(x\right)\left(x\ln\left(x\right)\left(\ln\left(x\right)+1\right)+1\right)+1\right) = x^{x^{x^x}+x^x-1}\overset{\small 2}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 That's nice! Let's try with xxxxxx^{x^{x^{x^x}}}: xxxxx+xxx1(xxxln(x)(xxln(x)(xln(x)(ln(x)+1)+1)+1)+1)=xxxxx+xxx1Bk=03kxln(x)β+1x^{x^{x^{x^x}}+x^{x^x}-1}\left(x^{x^x}\ln\left(x\right)\left(x^x\ln\left(x\right)\left(x\ln\left(x\right)\left(\ln\left(x\right)+1\right)+1\right)+1\right)+1\right) = x^{x^{x^{x^x}}+x^{x^x}-1}\overset{\small 3}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 Apart from the first bit it looks extremely similar to the previous one! To generalise the first bit, we can express it as xn1x+n2x1x^{ {^{n-1}x} + {^{n-2}x} - 1}

So finally, we can generalise the derivative of nx{^{n}x} as xn1x+n2x1Bk=0n2kxln(x)β+1x^{ {^{n-1}x} + {^{n-2}x} - 1}\overset{\small n-2}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 by using the new function.

I hope that this was not too confusing and that you may have learnt something!

(If you want the LaTeX\LaTeX for the nested bracket function, it is \overset{\small \text{some other number}}{\underset{\small k=\text{ a number}}{\huge \Beta }} ) Bk= a numbersome other number\overset{\small \text{some other number}}{\underset{\small k=\text{ a number}}{\huge \Beta }}

Note by James Watson
11 months ago

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