Cfrgtkky

I was recently watching blackpenredpen’s video (found here: https://m.youtube.com/watch?v=UJ3Ahpcvmf8) where he found the derivative of the the function $y = x^{x^x}$. Before watching the video, I decided to try it myself, and I didn’t it this way: Firstly I decided I would use the chain rule and define two functions, $u(x)$ and $v(x)$, both of which are $x^x$, so I could show the function as $u(v(x))$. Using the chain rule, the derivative would be: $u’(v(x))cdot v’(x)$. We can find the derivative of $x^x$ (as shown on Quora here: https://www.quora.com/What-is-the-derivative-of-x-x-How-could-I-derive-it-as-well) as $x^xcdot(ln x+1)$. Now using the chain rule to get $u’(v(x))$ we get ${(x^x)}^{(x^x)}cdot(ln {x^x}+1)$ or $x^{x^{x+1}}cdot(ln {x^x}+1)$. This wasn’t the answer that blackpenredpen arrived at though. In the video, he arrived at $x^xcdot x^{x^x}left({frac{1}{x}+ln x+(ln x)^2}right)$. I can’t see how he did it, and I’m sure that it is right. If anyone could confirm my answer or show where I went wrong it would be greatly appreciated.
Thanks.

You can apply the method given on quora to your problem:
$$left(x^{x^x}right)'=left(e^{x^xln x}right)'=e^{x^xln x}cdot left(x^xln xright)'=\
=e^{x^xln x}cdot left(e^{xln x}ln xright)'=\
=e^{x^xln x}cdot left[left(e^{xln x}right)'ln x+e^{xln x}cdot frac1xright]=\
=x^{x^x}cdot left[e^{xln x}left(xln xright)'ln x+x^xcdot frac1xright]=\
=x^{x^x}cdot left[x^xleft(ln x+1right)ln x+x^xcdot frac1xright]=\
=x^{x^x}cdot x^xleft[ln^2 x+ln x+frac1xright].$$
P.S. I could not watch the referenced video on Youtube, because it is not opening here (it might be blocked).

For differentiating expressions of the form $u(x)^{v(x)},$ I've always found it easiest to take logarithms and then differentiate implicitly. However, rather than work with the more general form $y = u^v,$ I'll work with $y = x^U$ (i.e. use $x$ for the base), which is sufficient to deal with $x^{x^x}.$

Taking the logarithm of both sides gives

$$ y = x^U ;; implies ln y = lnleft(x^U right) ;; implies ;; ln y = U cdot ln x $$

Now differentiate both sides with respect to $x$:

$$ frac{d}{dx}left(ln yright) ;; = ;; frac{d}{dx}left(U cdot ln x right) $$

$$ frac{y'}{y} ;; = ;; frac{d}{dx}(U) cdot ln x ; + ; U cdot frac{d}{dx} (ln x) ;; = ;; U'ln x ; + ; frac{U}{x} $$

$$ y' ;; = ;; y cdot left[ U'ln x ; + ; frac{U}{x} right] ;; = ;; x^U cdot left[ U'ln x ; + ; frac{U}{x} right] $$

Using this result for $U(x) = x,$ we get

$$ left(x^xright)' ;; = ;; x^x cdot left[ 1 cdot ln x ; + ; frac{x}{x} right] ;; = ;; x^x(ln x + 1) $$

Using that same result for $U(x) = x^x$ along with what we just found for $left(x^xright)',$ we get

$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ left(x^xright)' cdot ln x ; + ; frac{x^x}{x} right] $$

$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot left[ x^x(ln x + 1) cdot ln x ; + ; frac{x^x}{x} right] $$

$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x + 1)ln x ; + ; frac{1}{x} right] $$

$$ left(x^{x^x}right)' ;; = ;; x^{x^x} cdot x^x cdot left[ (ln x)^2 ; + ; ln x ; + ; frac{1}{x} right] $$