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

Jan 22, 2016

$\frac{d}{\mathrm{dx}} {x}^{{x}^{x}} = {x}^{{x}^{x} + x - 1} \left(x {\ln}^{2} \left(x\right) + x \ln \left(x\right) + 1\right)$

#### Explanation:

Use:

$\ln \left({a}^{b}\right) = b \ln \left(a\right)$

$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

Find:

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

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

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

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

So:

$\frac{d}{\mathrm{dx}} {x}^{{x}^{x}} = \frac{d}{\mathrm{dx}} {e}^{{e}^{x \ln \left(x\right)} \cdot \ln \left(x\right)}$

$= {e}^{{e}^{x \ln \left(x\right)} \cdot \ln \left(x\right)} \cdot \frac{d}{\mathrm{dx}} \left({e}^{x \ln \left(x\right)} \cdot \ln \left(x\right)\right)$

$= {x}^{{x}^{x}} \cdot \frac{d}{\mathrm{dx}} \left({e}^{x \ln \left(x\right)} \cdot \ln \left(x\right)\right)$

$= {x}^{{x}^{x}} \cdot \left(\left(\frac{d}{\mathrm{dx}} {e}^{x \ln \left(x\right)}\right) \cdot \ln \left(x\right) + {e}^{x \ln \left(x\right)} \cdot \left(\frac{d}{\mathrm{dx}} \ln \left(x\right)\right)\right)$

$= {x}^{{x}^{x}} \cdot \left({e}^{x \ln \left(x\right)} \left(\frac{d}{\mathrm{dx}} \left(x \ln \left(x\right)\right)\right) \cdot \ln \left(x\right) + {e}^{x \ln \left(x\right)} \cdot \frac{1}{x}\right)$

$= {x}^{{x}^{x}} \cdot {x}^{x} \cdot \left(\left(\frac{d}{\mathrm{dx}} \left(x \ln \left(x\right)\right)\right) \cdot \ln \left(x\right) + \frac{1}{x}\right)$

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

$= {x}^{{x}^{x}} \cdot {x}^{x} \cdot \left({\ln}^{2} \left(x\right) + \ln \left(x\right) + \frac{1}{x}\right)$

$= {x}^{{x}^{x} + x - 1} \left(x {\ln}^{2} \left(x\right) + x \ln \left(x\right) + 1\right)$