**Amherstclane**

Unregistered

Hey there,

First of all, I'm completely new to calculus in any form, so please excuse any completely obvious errors if there are any.

I was just wondering the other day about the derivative of a Power Tower. I tried to treat it as a x^n, and came up with

f(a, b) = a \uparrow \uparrow b

f'(a, b) = (a \uparrow \uparrow b) a^{(a \uparrow \uparrow [b-1]) -1}

Is this correct?

Cheers,

Amherst

Posts: 1,628

Threads: 105

Joined: Aug 2007

(11/14/2009, 05:50 AM)Amherstclane Wrote: Hey there,

First of all, I'm completely new to calculus in any form, so please excuse any completely obvious errors if there are any.

I was just wondering the other day about the derivative of a Power Tower. I tried to treat it as a \( x^n \), and came up with

\( f(a, b) = a \uparrow \uparrow b \)

\( f'(a, b) = (a \uparrow \uparrow b) a^{(a \uparrow \uparrow [b-1]) -1} \)

Is this correct?

Cheers,

Amherst

First a hint: you can enclose your formulas in tex tags like

gives the result \( x^n \). I did that for you in the quote.

And then for the derivative: If you have a function with two arguments you must specify with respect to which variable you are differentiating, e.g. \( \frac{\partial f(a,b)}{\partial a} \).

Before differentiating a whole powertower, let us start with something simpler: What is the derivative of \( x^x \)? You can not simply apply the \( \frac{\partial x^c}{\partial x} = c x^{c-1} \) rule, because this rule is only applicable for \( c \) being a constant.

If I slightly change the form and write \( x^x = e^{x \ln(x)} \) you can see that there are several functions involved of which you know the derivative already:

There is \( \exp'(x)=\exp(x) \), \( \ln'(x)=1/x \) and there is a product contained \( (f\cdot g)'(x) = f'(x) g(x) + f(x) g'(x) \), for nested functions you have the chain rule \( \frac{\partial f(g(x))}{\partial x}=f'(g(x)) g'(x) \).

So how can you apply all these rules to compute the derivative of \( x^x \)?