Why do the slopes of f(x)=x^x^x and f(x)=x^x^x^x behave unexpectedly?

[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!

 

[arildno]

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

 

[Reshma]

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

Good one!

 

[Office_Shredder]

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

 

[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.

 

[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.