# Differentiation question

1. May 6, 2005

### abia ubong

how can this be differentiated,i need to know how.x^x^x^x^x^x....
also ,my teacher posed this question to the whole class x^(x^2)^(x^3)^(x^4)................

2. May 6, 2005

### juvenal

The greatest mathematician of all time would be able to solve this easily.

3. May 6, 2005

### dextercioby

The first is really easy,if u consider logarithmic differentiation...

Daniel.

4. May 6, 2005

### kleinwolf

Try to write : h(x)=x^x^x^x....

Then clearly h(x)=x^h(x)....just differentiate and isolate h'(x)....

Also x^(x^2)=x^(x.x)=(x^x)^x=x^x^x...or if you want : log(x^(x^2))=x^2*log(x)=x*log(x^x)=log((x^x)^x)=log(x^x^x)

5. May 7, 2005

### abia ubong

you are really funny juvenal.
but then how about the other function, how can i get that differentiated?

6. May 7, 2005

### dextercioby

Kleinwolf,i'm sure you're not familiar to the notation $$x^{x^{x^{...}}}$$

So basically i could write

$$x^{x^{2}}=x^{x\cdot x}=\left(x^{x}\right)^{x}\neq x^{x^{x}}}$$

Daniel.

7. May 8, 2005

### extreme_machinations

hey if its x^x^x^.......infinity
then
y = x^x^x^x^......can be written as

y = x^y ,
then take log of both sides and then diffrentiate .

8. May 8, 2005

### kleinwolf

Right Daniel, I messed up...

I got disturbed by having no parenthesis, because $$x^{(x^2)}\neq (x^x)^2=x^{(2x)}$$.....

Then I think u could use an iterative definition : $$f(x,y)=x^y$$

$$h_1(x)=x, h_2(x)=x^{(x^2)}, h_3(x)=x^{((x^2)^{(x^3)})}...$$

So that in fact : $$h_1(x)=f(x,1), h_2(x)=f(x,x^2), h_3(x)=f(x,f(x^2,x^3))...aso...$$

Now you "just" have to differentiate the imbrication of f(...f(...(x^{n-1},x^n)...)....this should be quite complicated.

Last edited: May 8, 2005
9. May 8, 2005

### abia ubong

how about the other question ,no one is talking about that ,just the easy one.
the most important is x^(x^2)^(x^3)....

10. May 9, 2005

### uart

For the first one I got the following answer, which I'm pretty sure is correct.
y' = y^2 / ( x ( 1 - ln(x) y ) )

I set it up as an implicit function, y(x) : (x^y - y)=0, and used
y' = -(dQ/dx) / (dQ/dy), where Q(x,y) = x^y - y.

Last edited: May 9, 2005
11. May 9, 2005

### uart

Recalling the ambiguity (that was clarified by Daniel) in the first expression is that second one meant to be evaluated from left to right or from right to left ?

If it's evaluated left to right then it diverges for all x>1 and is one for 0<x<=1, so the derivative is not really very interesting as it's just zero in the region for which the function is meaningfully defined. If however it's evaluated right to left then hmmm, I'm not sure.

12. May 10, 2005

### mathelord

That is the solution for the first function ,how about the other x^(x^2)^(x^3)^(x^4).....................