## Calculus 10th Edition

$2x\ln x+x$
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: $y^{\prime}=(x^{2})^{\prime}\cdot\ln x+x^{2}\cdot(\ln x)^{\prime}$ $=2x\displaystyle \ln x+x^{2}\cdot\frac{1}{x}$ $=2x\ln x+x$