Calculating $\frac{f'(e)}{f(e)}$ for $f(x) = x^{x^{x}}$.

Question

Define $f(x) = x^{x^{x}}$

Compute $\frac{f'(e)}{f(e)}$.

I have an answer, but I do not know how to reach this answer. The correct answer is: $e^{e-1}+2e^e$

Answer 1

To start with $$ \frac{f'(x)}{f(x)} = \frac{d}{dx}\ln f(x) $$ Since $\ln f(x) = x^x \ln(x)$, the product rule gives $$ \frac{f'(x)}{f(x)} =\frac{d(x^x)}{dx}\ln x + \frac{x^x}{x} $$ We can use $x^x = e^{x\ln x}$ to get $d(x^x)/dx = x^x(1+\ln x)$, which brings us to $$ \frac{f'(x)}{f(x)} =x^x(1+\ln x)\ln x + x^{x-1} $$ Plugging in $e$ then gives your answer.

Answer 2

We let $f (x)=x^{x^{x}} $. Then we have $$f'(x)=x^{x^{x}} \frac {d}{dx}[x^x \ln x] $$ $$=x^{x^{x}} [ \frac{d}{dx}(x^x)\ln x +x^x \frac {d}{dx}(\ln x)] $$ $$=x^{x^{x}}[x^x \ln x(\ln x+1)+x^{x-1}] $$ Hope you can take it from here.

Answer 3

$ f(x) = x^{x^x} \\ \implies \ln f = x^x \ln x \\ \implies \frac{d}{dx} \ln f = \ln x \frac{d}{dx} (x^x) + (x^x) \frac{d}{dx} (\ln x) \\ \implies \frac{f'(x)}{f(x)} = x^x \ln x (1 + \ln x) + x^{x - 1}. $