How do you find the derivative of x^lnx?

Jun 25, 2016

Answer:

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

Explanation:

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

hence $\ln y = \ln x \times \ln x = {\left(\ln x\right)}^{2}$

hence $\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln x \times \frac{1}{x}$

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