# Tricky derivation...

• Jan 24th 2007, 05:20 AM
kukkeball
Tricky derivation...
Can someone help me to derivate x^x^x? I'm starting to belive it's not possible... : D heeelp! :) Thank you in advance!
• Jan 24th 2007, 05:42 AM
earboth
Quote:

Originally Posted by kukkeball
Can someone help me to derivate x^x^x? I'm starting to belive it's not possible... : D heeelp! :) Thank you in advance!

Hello,

you only have to change the x's into powers of base e:

$\displaystyle x^{x^x}=e^{e^{x \cdot \ln(x)}}$

I presume that you know the chain rule and that you know how to derivate an exponentila function with base e.

EB
• Jan 24th 2007, 05:49 AM
AfterShock
Quote:

Originally Posted by kukkeball
Can someone help me to derivate x^x^x? I'm starting to belive it's not possible... : D heeelp! :) Thank you in advance!

Oh, it's perfectly possible, and fun too ;) .

I have class soon, but I will get you started.

First, what is the derivative of x^x?

We have y = x^x

Note: x > 0

ln(y) = ln(x^x)

ln(y) = x*(ln(x))

Differentiate both sides with respect to x;

y'(1/y) = ln(x) + x*(1/x) = (ln(x) + 1)

y' = (ln(x) + 1)*y

But recall y = x^x

y' = (ln(x) + 1)*x^x

Now you can use this to find the derivative of x^x^x.
• Jan 24th 2007, 06:13 AM
AfterShock
Quote:

Originally Posted by earboth
Hello,

you only have to change the x's into powers of base e:

$\displaystyle x^{x^x}=e^{e^{x \cdot \ln(x)}}$

I presume that you know the chain rule and that you know how to derivate an exponentila function with base e.

EB

Perhaps I am missing something, but how is e^e^(x*ln(x)) = e^x^x the same as x^x^x. The derivatives are clearly different.
• Jan 24th 2007, 10:48 AM
earboth
Quote:

Originally Posted by AfterShock
Perhaps I am missing something, but how is e^e^(x*ln(x)) = e^x^x the same as x^x^x. The derivatives are clearly different.

Hello, AfterShock,

of course you are right: I messed up my notes. It should be:

$\displaystyle x^{x^x}=e^{\ln(x) \cdot e^{x*ln(x)}}$

As you can see, one factor ln(x) slipped away. But anyhow your method is much better as what I've planned to do.

EB
• Jan 25th 2007, 08:54 AM
kukkeball
So what's the correct answer... i got something crazy... : F
• Jan 25th 2007, 11:34 AM
Soroban
Hello, kukkeball!

Here's another approach . . .

Quote:

Differentiate: . $\displaystyle y \:=\:x^{x^x}$

Take logs: .$\displaystyle \ln y \;=\;\ln\left(x^{x^x}\right) \;=\;x^x\ln x$

Take logs again: .$\displaystyle \ln(\ln y)\;=\;\ln\left[x^x\ln x\right] \;=\;\ln\left(x^x\right) + \ln(\ln x) \;=\;x\ln x + \ln(\ln x)$

Differentiate implicitly (very carefully) . . .

. . $\displaystyle \frac{1}{\ln y}\cdot\frac{1}{y}\cdot\frac{dy}{dx}\;=\;x\cdot\fr ac{1}{x} + \ln x + \frac{1}{\ln x}\cdot\frac{1}{x}$

We have: .$\displaystyle \frac{1}{y\ln y}\cdot\frac{dy}{dx} \;=\;1 + \ln x + \frac{1}{x\ln x}$

Then: .$\displaystyle \frac{dy}{dx}\;=\;y\ln y\left(1 + \ln x + \frac{1}{x\ln x}\right)$

Therefore: .$\displaystyle \frac{dy}{dx}\;=\;x^{x^x}\ln\left(x^{x^x}\right)\l eft(1 + \ln x + \frac{1}{x\ln x}\right)$