# How do you differentiate (lnx)^(x)?

Dec 24, 2016

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

#### Explanation:

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

Take (Natural) logarithms of both sided:

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

Differentiate Implicitly (LHS) and apply product rule and chain rule (RHS).

 \ \ \ \ \ \ 1/ydy/dx = (x)(1/lnx*1/x) + (1)(ln((lnx))
$\therefore \setminus \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} x + \ln \left(\left(\ln x\right)\right)$
$\therefore \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = y \left\{\frac{1}{\ln} x + \ln \left(\left(\ln x\right)\right)\right\}$
$\therefore \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\ln x\right)}^{x} \left\{\frac{1}{\ln} x + \ln \left(\left(\ln x\right)\right)\right\}$