# How do you differentiate x^lnx?

Nov 21, 2016

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

#### Explanation:

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

Take Natural logarithms of both sides:
$\ln y = \ln \left\{{x}^{\ln x}\right\}$
 :. lny = (lnx)(ln{x)  as $\ln {a}^{b} = b \ln a$
$\therefore \ln y = {\left(\ln x\right)}^{2}$

Differentiate Implicitly (LHS), and apply chain rule to RHS:
$\therefore \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(\ln x\right) \left(\frac{1}{x}\right)$
$\therefore \frac{1}{{x}^{\ln x}} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(\ln x\right) {x}^{-} 1$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left({x}^{\ln x}\right) 2 \left(\ln x\right) {x}^{-} 1$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{\ln \left(x\right) - 1} \ln x$