Weird function!

1. Nov 6, 2006

Rishav

Hey people,
Consider the function:
f(x)=x^x^x
At 0,f(x)=0 and d/dx=1
At 1,f(x)=1 and d/dx=1
At 2,f(x)=16 and d/dx=107.11
At 3,f(x)=7.625597485000*10^12 and d/dx=5.43324993100000*10^14

Why this??? Check the slopes...

And: Slopes between 0-1
At 0.1, f(x)=0.16057 d/dx= 1.658
At 0.2, f(x)=0.31146 d/dx=1.350
At 0.3, f(x)=0.43215 d/dx=1.070
At 0.4, f(x)=0.52987 d/dx=0.890
At 0.5, f(x)=0.61255 d/dx=0.777
At 0.6, f(x)=0.68662 d/dx=0.716
At 0.7, f(x)=0.75740 d/dx=0.708
At 0.8, f(x)=0.82972 d/dx=0.747
At 0.9, f(x)=0.90862 d/dx=0.840
At 1.0, f(x)=1.00000 d/dx=1.000
The slopes between 0.2-0.7 are decreasing...why's that??? Can anyone explain?

And one more function f(x)=x^x^x^x
At 0, d/dx=-infinity
At 0.1, d/dx=-1.528330000
At 0.2, d/dx=-0.37292
At 0.3 d/dx~0
At 0.4 d/dx=0.31333
At 0.5 d/dx=0.45030
At 0.6 d/dx=0.54327
At 0.7 d/dx=0.63323
At 0.8 d/dx=0.72329
At 0.9 d/dx=0.83695
At 1.0 d/dx=1.000

So can anyone say why does the slope decrease and then increase so rapidly??? Hope my questions are not falling on deaf ears!

2. Nov 6, 2006

arildno

Well, you might try to differentiate the animal and verify your observations!

3. Nov 6, 2006

Reshma

:rofl: :rofl: Good one!

4. Nov 6, 2006

Office_Shredder

Staff Emeritus
Is this ((x^x)^x)^x or (what I'm assuming) x^(x^(x^x))?

EDIT: Or I can just be smart and plug in one of the values you've so kindly given us. Never mind

5. Nov 6, 2006

arildno

It isn't that difficult to differentiate the creature.
Let's tackle the first one:
$$f(x)=x^{g(x)}=e^{\ln(x)g(x)}, g(x)=x^{x}=e^{x\ln(x)}$$
Thus,
$$\frac{df}{dx}=f(x)*(\frac{g(x)}{x}+\ln(x)g'(x))=f(x)*(\frac{g(x)}{x}+\ln(x)g(x)(\ln(x)+1))$$

This might then be written out more fully, if that's important to you.

6. Nov 6, 2006

ObsessiveMathsFreak

Just plot the function, alongside x^x. The reason the slopes decrease is beacaues x^x is a decreasing function for a short time in and around that range.