Derivative of x{^{\infty}x}??? (Part 3 of my past 2 notes)

My last note went over differentiating nx{^{n}x} and it turned out to be 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 where Bk=abf(x)βg(x)=f(x)(f(x)(f(x)g(x))g(x)))g(x)ba nested parentheses\overset{\small b}{\underset{\small k=a}{\huge \Beta }}f(x)\beta g(x) = \underbrace{f(x)(f(x)(\dots f(x)g(x))g(x))\dots)g(x)}_{b-a \text{ nested parentheses}} However, I feel that there is one more problem for this trilogy of notes to acknowledge: the derivative of x{^{\infty}x}. We can utilise our generalisation from earlier to solve this: ddx(x)=x1x+2x1Bk=02kxln(x)β+1=xx+x1Bk=0kxln(x)β+1=x2(x)1Bk=0kxln(x)β+1\begin{aligned} \frac{d}{dx}({^{\infty}x}) &= x^{ {^{\infty-1}x} + {^{\infty-2}x} - 1}\overset{\small \infty-2}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 \\ &= x^{ {^{\infty}x} + {^{\infty}x} - 1}\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 \\ &= x^{ 2({^{\infty}x}) - 1}\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 \end{aligned} But what is Bk=0kxln(x)β+1\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1? It is the same as xln(x)(1xln(x)((ln(x)+1)+1))+1infinite nested parentheses\underbrace{ {^{\infty}x}\ln(x)({^{\infty-1}x}\ln(x)( \dots( \ln(x) +1)+1)\dots)+1}_{\text{infinite nested parentheses}} but can this be expressed more clearly? Maybe. Let us have another go at differentiatingx{^{\infty}x}, but using a different method.

The equation xy=yx^y = y is the same as x=y{^{\infty}x} = y. This is because y=xy=xxy=xxxy=xxxx=xy = \blue{x^y} = x^{\blue{x^y}} = x^{x^{\blue{x^y}}} = x^{x^{x^{x^{\dots}}}} = {^{\infty}x}. We can use implicit differentiation here to find the derivative of x{^{\infty}x}: ddx(y=xy)    dydx=xyddx(yln(x))=xy(ln(x)dydx+yx)=xyln(x)dydx+xyyx\begin{aligned} \frac{d}{dx}\left( y = x^y \right) \implies \frac{dy}{dx} = x^y \cdot \frac{d}{dx}(y\ln(x)) &= x^y \left( \ln(x)\frac{dy}{dx} + \frac{y}{x} \right)\\ &= x^y \ln(x)\frac{dy}{dx} + \frac{x^y y}{x} \end{aligned}     dydxxyln(x)dydx=xyyx\implies \frac{dy}{dx} - x^y \ln(x)\frac{dy}{dx} = \frac{x^y y}{x}     dydx(1xyln(x))=xy1y\implies \frac{dy}{dx}(1-x^y\ln(x)) = x^{y-1} y     dydx=xy1y1xyln(x)\implies \frac{dy}{dx} = \boxed{\frac{x^{y-1} y}{1-x^y\ln(x)}} Note that y=xyy = x^y so we can also write this as xy1xy1xyln(x)=x2y111xyln(x)\frac{x^{y-1}x^y}{1-x^y\ln(x)} = x^{2y-1}\frac{1}{1-x^y\ln(x)} Also remember that y=xy = {^{\infty}x}, so it can be written as x2(x)111xln(x)x^{2({^{\infty}x})-1}\frac{1}{1-{^{\infty}x}\ln(x)} This must be the same as our first attempt on the derivative of x{^{\infty}x}, so we may be able to figure out what the infinite nested brackets could be: x2(x)111xln(x)=x2(x)1Bk=0kxln(x)β+1\blue{x^{2({^{\infty}x})-1}}\frac{1}{1-{^{\infty}x}\ln(x)} = \blue{x^{ 2({^{\infty}x}) - 1}}\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1     Bk=0kxln(x)β+1=11xln(x)\implies \green{\boxed{\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 = \frac{1}{1-{^{\infty}x}\ln(x)}}} So the derivative of x{^{\infty}x} is x2(x)111xln(x)x^{2({^{\infty}x})-1}\cfrac{1}{1-{^{\infty}x}\ln(x)}, but we also figured out that Bk=0kxln(x)β+1=11xln(x)\overset{\small \infty}{\underset{\small k=0}{\huge \Beta }}{^{k}x}\ln(x)\beta + 1 = \cfrac{1}{1-{^{\infty}x}\ln(x)} which is quite cool if i do say so myself.

I hope that you found this interesting!

Note by James Watson
3 weeks, 3 days ago

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