## Calculus: Early Transcendentals 8th Edition

$y'=x^{x}(1+lnx)$
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, $y'=x^{x}(1+lnx)$