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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
(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
Code:
$$x^n$$
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 $x^x$?