# How do you differentiate y=x^x?

Mar 5, 2015

You can use logarithmic differentiation

Take the natural logarithm of both sides

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

Now using properties of logarithms, rewrite the right hand side

$\ln y = x \ln x$

Differentiate both sides with respect to $x$
Use the product rule on the right side

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \ln x + x \frac{1}{x}$

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \ln x + 1$

Multiply both sides by $y$

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

Now $y = {x}^{x}$ so we can write

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