Answer
$y'=x^{x}(1+lnx)$
Work Step by Step
Given: $y =x^{x}$
Taking logarithmic on both sides of the function$y =x^{x}$.
Use logarithmic property $ln(x^{y})=ylnx$
$lny=xlnx$
Take implicit differentiation with respect to $x$.
Apply product rule of differentiation.
$\frac{d}{dx}(lny)=\frac{d}{dx}(xlnx)$
$\frac{1}{y}\frac{d}{dx}(y)=x\frac{d}{dx}(lnx)+lnx\frac{d}{dx}(x)$
$\frac{d}{dx}(y)=y[x\times(\frac{1}{x})+lnx\times1]$
Hence, $y'=x^{x}(1+lnx)$
