I'm very interesting in functions of the nature:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]f(x) = x^{x}[/tex]

[tex]f(x) = x^{x^{x}}[/tex]

and so on. I believe these are called tetrations? Regardless, I sought to generalize the nth derivative of [itex]f(x)=x^x[/itex] and it is proving to be difficult.

First I tried just repeatedly differentiating until I could see a pattern:

[tex] f'(x) = x^{x}\left(1+\ln(x)\right)[/tex]

[tex]f''(x) = x^{x-1}\left(1+x+2x\ln(x) + x\ln^{x}(x)\right) [/tex]

[tex]f'''(x) = x^{x-2}\left( -1+3x+x^{2} +3x(1+x)\ln(x)+3x^{2}\ln^{2}(x)+x^{2}\ln^{3}(x)\right) [/tex]

[tex]f^{4}(x) = x^{x-3}\left( 2-x+6x^{2}+x^{3}+4x(-1+3x+x^{2})\ln(x)+6x^{2}(1+x)\ln^{2}(x)+4x^{3}\ln^{3}(x)+x^{3}\ln^{4}(x) \right) [/tex]

I couldn't seem to find a pattern here. So I tried representing [itex] f(x)=x^{x} [/itex] as a power series:

[tex] f(x) = x^{x} = e^{x\ln(x)} = \sum_{n=0}^{\infty} \frac{1}{n!}x^n\ln^{n}(x) [/tex]

Then maybe taking the derivatives of the power series could lead me to an easier pattern. After I took the first derivative I felt that it wasn't the case as:

[tex] f'(x) = \sum_{n=0}^{\infty} \frac{1}{(n-1)!}\left(1+\ln(x)\right)x^{n-1}\ln^{n-1}(x) [/tex]

Which looks to be at a first glance more complex. Putting it through mathematica kind of left me with the same feeling of getting no where.

Is there anything I should be considering? Any methods that I could employ to solve my problem? I don't know much about higher level maths (I'm currently doing Calc III) .

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# I General Nth Derivative

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