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Trouble to find answer - Amherstclane - 11/14/2009
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
Derivative of powertower - bo198214 - 11/14/2009
(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:
[tex]x^n[/tex]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 \( x^x \)?