## Calculus, 10th Edition (Anton)

$dy/dx=\ln(x)+1$
Using the product rule for derivatives ($[uv]'=u'v+uv'$), the derivative of $xln(x)$ equals $(x)'ln(x)+x(ln(x))'$=$ln(x)+\frac{x}{x}=\boxed{\ln(x)+1}$.