1. ## 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!

2. 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

3. 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.

4. 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.

5. 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

6. So what's the correct answer... i got something crazy... : F

7. Hello, kukkeball!

Here's another approach . . .

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