Answer
$2x\ln x+x$
Work Step by Step
By theorem 5.3/$1$:$\ \ \ \displaystyle \frac{d}{dx}[\ln x]=\frac{1}{x}$
$y=x^{2}\cdot\ln x$
We use the product rule here:
$y^{\prime}=(x^{2})^{\prime}\cdot\ln x+x^{2}\cdot(\ln x)^{\prime}$
$=2x\displaystyle \ln x+x^{2}\cdot\frac{1}{x}$
$=2x\ln x+x$
