derivative of tetration of x (hyperpower)

Derivative of tetration of x, derivative of (x↑↑3), derivative of double up arrows, derivative of tetration of x, Knuth Arrow Notation. This is a hard calculus 1 problem when we differentiate this power tower of x

#blackpenredpen #calculus #math #tutorial #college

[Lagiacrusguy]

“i don’t know how to integrate this so don’t ask me”

we found his kryptonite

[coldlogiccrusader]

IU am 66 yrs old. I earned a MS in Mathematical Physics in 1977. I never heard about Tetration till just now THANK YOU so

[sreekommalapati]

"Im just gonna put this in the thumbnail to make a little clickbait"

Transparency

[tbonbt]

For those who'd like to do more research on ⁿa, the notation is called tetration.

[Sid-ixqr]

If you'd put d/dx (x³) for my final exam, you'd be my favourite teacher!

[helio]

X: This isn't even my final form!

[TurdFurgeson]

Tetration, not to be confused with titration. I wonder how many chemistry students came here looking for curves and then subsequently ran away because of some light mathing. Love the video. Clear and concise. It's clear that your professor chops are strong.

[MegaPhester]

I found a nice recursive formula for the derivative of x↑↑n by setting y = x↑↑n, taking the log on both sides and doing implicit differentiation:
d(x↑↑n) = x↑↑n * ( d(x↑↑(n-1)) * log(x) + x↑↑(n-1) / x )

[heliocentric]

"Chen Lu" The Goddess of Derivatives

[silasrodrigues]

I have never seen this notation before!

[valeriebarker]

For any function f(x)^g(x), the derivative can be found by adding the power rule to the exponent rule. That is to say d/dx (f(x)^g(x)) =

[suleem]

You taught me so many things like double factorials, hyperpowers... I never thought such things exist. Well done!

[gamingletsplays]

It's so amazing to see a (mostly) friendly community of people who like math as much as me (:

[andrewcorrie]

Love your videos: I haven't studied Maths for a long time, and neither do I teach it, but these make difficult problems so easy to follow.

[fmakofmako]

I didn't read through the comments so if someone has already posted the derivative then kudos to them.

The form of the derivative of x^^n for n >=4 is:



You can prove this by induction. The inductive step is shown by the recursion derivative of x^^n = and the base case is that the derivative of x^^4 is

I put the base case in the same form as my answer to show that it's true because yt comments are hard to format and anyways loads of people in the comments did the x^^4 case.

Moving on to the induction we have from the recursion. Then we plug in the same form from above.



Therefore:



Done.

[novidsonmychanneljustcomme]

In the ending result you also could simplify: x^(x^x)*x^x = x^(x^x+x), but this depends on which notation you prefer. ;-)

[kruksog]

Hey BPRP. I'm about to graduate with a math degree. I want to tell you, your vids are awesome and I love your pronunciation. Thanks for what you do! You bring me snippets of math I can enjoy when I feel bogged down in technical math i have to learn for a grade (which I'm sure you know can suck the fun out of it). So thanks. Truly.

[zeldasama]

My mans ADMITTED he was gonna put tetration of x in video for a clickbait. HahAHA

[jamez]

For pentation, you have to use the up arrows, presumably because you run out of upper corners to write in after tetration due to the number of upper corners being 2...

[intergalakti]

Hi, I really like your videos and since you used tetration in this one, I would like to ask you a question that's been on my mind for quite some time now, but for which I could not find a solution yet.
What I realized was the following: If you try to complete the natural numbers with respect to subtraction, which is the inverse operation to multiplication, you get the integers.
If you complete these with respect to quotients, which are inverse to multiplication, i.e. form the quotient field, you obtain the rational numbers.
By completing these wrt. roots of polynomials, i.e. wrt. exponentiation, you obtain the algebraic numbers.
But what if you complete these using tetration, i.e. add superroots, superlogs and other solutions of "tetration equations"? E.g. the superroot of 2, i.e. the number x with x^x=2, seems to not be algebraic, so can you form a field extending the rational numbers that is closed wrt. these operations? It can't be a field extension of finite dimension since it is not algebraic, and it also has to be a field, I think, and it should still be countable so it isn't the real numbers, but I could not figure out much more.
Also, if you continue this process with pentation and higher order operations (see Knuth's up-arrow notation), you should get other fields extending each other. Since they are a mapping family, you can take the direct limit of those, and get a very big field - are these the whole real numbers? A lot of questions, I know, but I hope someone else already thought about it and figured it out, since it is far beyond my current scope. Anyway, I would be happy about any kind of help.