# What is the derivative of x^x^x?

##### 1 Answer
May 22, 2015

Since the rule:

${\left[f \left(x\right)\right]}^{g} \left(x\right) = {e}^{\log} \left({\left[f \left(x\right)\right]}^{g} \left(x\right)\right) = {e}^{g \left(x\right) \log f \left(x\right)}$,

then:

${x}^{{x}^{x}} = {e}^{\log} \left({x}^{{x}^{x}}\right) = {e}^{{x}^{x} \log x} = {e}^{{e}^{\log} \left({x}^{x}\right) \log x} =$

$= {e}^{\left({e}^{x \log x}\right) \log x}$.

So the function to derive is:

$y = {e}^{\left({e}^{x \log x}\right) \log x}$.

$y ' = {e}^{\left({e}^{x \log x}\right) \log x} \cdot \left[{e}^{x \log x} \left(1 \cdot \log x + x \cdot \frac{1}{x}\right) \cdot \log x + {e}^{x \log x} \cdot \frac{1}{x}\right] =$

$= {e}^{\left({e}^{x \log x}\right) \log x} \cdot {e}^{x \log x} \left(\log x + 1 + \frac{1}{x}\right) =$

$= {e}^{\left({e}^{x \log x}\right) \log x} \cdot {e}^{x \log x} \frac{x \log x + x + 1}{x}$

And, if you want:

$y ' = {x}^{{x}^{x}} \cdot {x}^{x} \frac{x \log x + x + 1}{x}$.