Taking the derivative of $f(x) = x^{x^x}$?

Question

Let $f(x) = x^{x^x}$. I checked the derivative of this function using Wolfram Alpha.

I know that we can find the derivative of this by using $$ f'(x) = e^{x^x \log(x)}, $$ to get $f'(x) = x^{x^x+x-1}(x\log^2(x) + x \log(x) + 1)$, which is the correct answer.

But if we instead use $$ f'(x) = e^{x \log(x^x)} $$ we get the wrong answer $f'(x) = (x^x)^x(\log(x^x) + x + x\log(x)).$

Why do we get the wrong answer using the second approach, it seems bringing the $x$ exponent out of the log should be equivalent to bringing the $x^x$ exponent out of the log..but for some reason it isn't. Why?

Answer 1

Note that $\log(x^x)=x\log x$, so $x\log(x^x)=x^2\log x\not=x^x\log x$ (except, of course, when $x=1$ or $2$). It follows that, as functions, $e^{x\log(x^x)}\not=e^{x^x\log x}$.

Answer 2

$e^{x\log(x^x)} = (x^x)^x$, whereas $e^{x^x\log(x)} = x^{(x^x)}$.

Answer 3

$$u=x^x \implies \ln u = x \ln x \implies \frac{u'}{u}=x\frac{1}{x}+\ln x \implies u'= x^x(1+\ln x).$$ Next $$v=x^{x^x} \implies \ln v = u \ln x \implies \frac{v'}{v}=u' \ln x+u \frac{1}{x}$$ $$\implies v'= v u' \ln x+\frac{uv}{x}=x^{x^x}~x^x~ [(1+\ln x)~\ln x+\frac{1}{x} ]$$

Answer 4

Just to clarify some of the other answers with a plot. Part of the confusion may stem from the fact that

$$f(x) = {(x^x)}^x\\g(x) = x^{(x^x)}$$

are, in fact, different functions. This becomes particularly clear around $x=1.7$ below.