Let's try this! @ikram002p
\[\frac{d}{dx}(a\uparrow^x b)\]

- Kainui

Let's try this! @ikram002p
\[\frac{d}{dx}(a\uparrow^x b)\]

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- katieb

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- Here_to_Help15

Can i try as well?

- Kainui

Yeah anyone can lol

- Here_to_Help15

Oki lol let me get paper

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## More answers

- Kainui

This might be an impossible question lol.

- Here_to_Help15

lol hmm nothing is impossible ;)

- ikram002p

i'll write what im thinking of and then lets see :)

- anonymous

Woah let me take a stab at this!

- Here_to_Help15

Can i give you a question @Kainui :)

- Here_to_Help15

\[\frac{ d }{ dx } (2^{x)}\]
How do you find ^

- Here_to_Help15

You lost @Kainui ? tehee :)

- ikram002p

@Kainui
lets try on small stuff like
|dw:1433362211420:dw|

- ganeshie8

* for tomoro

- Here_to_Help15

?

- ikram002p

|dw:1433362587173:dw|

- Here_to_Help15

Is that

- Here_to_Help15

Your writing

- ikram002p

thats only special case
yes my writing

- ikram002p

eh -.- i wish if there is a short cut

- Here_to_Help15

lol neat hand writing :)

- ikram002p

ty :)

- ikram002p

i might end up saying zero since this arrow notation end up defining a finite number hmmm
but if we wanna deal with it as exponent we neat ani_arrow notation , right @Kainui ? is there something like this ??

- Kainui

Sorry I got distracted right as I posted this orry!
Good idea on using the 2 case, that's awesome!

- Kainui

One way I was thinking of is seeing if we could weasel our way into a continuous definition of the arrow notation with the definition of derivative:
\[\large \lim_{h \to 0}\frac{a \uparrow^{x+h} b - a \uparrow^x b }{h}\]

- Kainui

For anyone who doesn't know the arrow notation, which to be honest I barely do either. \[ a \uparrow b = a^b \\ a \uparrow \uparrow b = a^{a^{a^\cdots}} \text{tower b high}\] So the recursive definition is found here, but I'llt ype it out too:
\[a \uparrow ^n b = a \uparrow ^{n-1}[a \uparrow^n(b-1)]\]
and
\[a \uparrow^n 1 = a\]
http://mathworld.wolfram.com/PowerTower.html
So for example: \[4 \uparrow^2 3 = 4 \uparrow (4 \uparrow^2 2)=4 \uparrow (4 \uparrow (4 \uparrow^2 1)) = 4 \uparrow (4 \uparrow (4 )) = 4^{4^4}\]

- geerky42

I don't know, but I think we may need to look at how \(\uparrow^x\) is defined for any real value of x?

- Kainui

Yeah it relies on making this definition which is discrete into something continuous which might not really even be possible.

- Kainui

Here I should have posted this to begin with for those who aren't into this yet!
http://www.numberphile.com/videos/grahamsnumber.html

- Kainui

Another example, here are our rules just for reference:
\[a \uparrow ^n b = a \uparrow ^{n-1}[a \uparrow^n(b-1)]\]
\[a \uparrow^n 1 = a\]
Ok so the example: \[4 \uparrow^3 3 \] Just plug and chug that recursion relation:\[ 4 \uparrow^2 (4 \uparrow^3 2) \]Now just looking at that part \((4 \uparrow^3 2)\) I expand that further:\[ 4 \uparrow^2 (4 \uparrow^2(4 \uparrow^3 1)) \]
now using the fact that \((4 \uparrow^n 1) = 4 \) also listed above as the other rule we have:
\[ 4 \uparrow^2 (4 \uparrow^24) \]
Which we then expand further

- Kainui

The next step we start expanding:
\[ 4 \uparrow^2 (4 \uparrow 4 \uparrow 4 \uparrow 4) = 4 \uparrow^2 4^{4^{4^4}}\]
And past this we really can't do anything in terms of writing it cause this is a tower of 4s that is \(4^{4^{4^4}}\) high. lol

- geerky42

Here's another problem for you to consider into:
Evaluate \(\dfrac{\mathrm d^{1/2}}{\mathrm dx^{1/2}}~~\large x\)
:)

- ikram002p

a way not impossible

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