Let $y=x^{x}$
Taking natural log on both the sides, we get
$\ln\ y=x\ln\ x$
Differentiate both the sides wrt to x we get
$\frac{1}{y}\frac{\mathrm{d}y }{\mathrm{d} x}$ = $(1) \ln \ x +x\frac{1}{x} $ (Product Rule)
$\frac{\mathrm{d}y}{\mathrm{d}x}$ = $y \ln\ x + y$
Substitute $y = x^{x}$
$\frac{\mathrm{d}y }{\mathrm{d} x}$=$\ x^{x}\ln\ x + x^{x}$
Hence, option C is correct answer.