# How do you differentiate f(x)=ln(x)^x?

Jan 20, 2016

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

#### Explanation:

Using implicit differentiation , the chain rule , and the product rule ,

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

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

$\implies \frac{d}{\mathrm{dx}} \ln \left(y\right) = \frac{d}{\mathrm{dx}} x \ln \left(\ln \left(x\right)\right)$

$\implies \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = x \left(\frac{d}{\mathrm{dx}} \ln \left(\ln \left(x\right)\right)\right) + \ln \left(\ln \left(x\right)\right) \left(\frac{d}{\mathrm{dx}} x\right)$

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

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

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

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\frac{1}{\ln} \left(x\right) + \ln \left(\ln \left(x\right)\right)\right)$

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