Find derivative of $x^{x^x}$

Question

Trying to find the derivative of: $$ x^{x^x} $$

I have a solution but cannot understand the third transition:

Answer 1

Hint: Use the product rule

$$(\ln x e^{x \ln x})' = \frac{1}{x}e^{\ln x^x} + \ln x(\ln x + 1)e^{ \ln x^x} = \frac{1}{x} x^x + \ln x(\ln x + 1) x^x$$

Answer 2

$$ x^{x^x} = x^y = e^{y\ln x} $$

Then the derivative in terms of $y$ is $$ \frac{d}{dx}e^{y\ln x} = e^{y\ln x}\left( y'\ln x + \frac{y}{x}\right) $$ So now you can compute $y'$ right? Using a similar process.

A quick tip, Always break the problem down into steps when you are learning new techniques :).

Answer 3

Generally ,if f(x) and g(x) are functions than $(u^{v})´=vu^{v-1}u´+u^{v}v´\ln u$

Answer 4

If $f(x)=x^{x^{\scriptstyle x}}$ and $g(x)=x^x$, then $$ \log f(x)=x^x\log x=g(x)\log x $$ so $$ \frac{f'(x)}{f(x)}=g'(x)\log x+\frac{g(x)}{x} $$ Now $$ \log g(x)=x\log x $$ so $$ \frac{g'(x)}{g(x)}=\log x+1 $$ Putting things together \begin{align} f'(x)&=f(x)\left(g(x)(\log x+1)\log x+\frac{g(x)}{x}\right)\\ &=x^{x^{\scriptstyle x}}x^x\left((\log x+1)\log x+\frac{1}{x}\right) \end{align}