Differentiate y=x^x?

Mar 27, 2016

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

Explanation:

As $y = {x}^{x}$, we have $\ln y = \ln {x}^{x} = x \ln x$

Hence differentiating both sides, $\frac{1}{y} \times \frac{\mathrm{dy}}{\mathrm{dx}} = 1 \times \ln x + x \times \frac{1}{x}$

or $\frac{1}{{x}^{x}} \times \frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \ln x$ and hence

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