## Calculus (3rd Edition)

$$y' =x(\ln x^2 + 1)$$
Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\ln x)'=\dfrac{1}{x}$ Recall that $(x^n)'=nx^{n-1}$ Since $y=x^2 \ln x$, then using the product rule we have $$y'=2x \ln x+ x^2 \frac{1}{x}=2x \ln x +x=x(\ln x^2 + 1)$$