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 Trouble to find answer Amherstclane Unregistered   11/14/2009, 05:50 AM 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 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 11/14/2009, 08:43 AM (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 $x^x$? « Next Oldest | Next Newest »

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