Product rule with a logarithm

Identify the two factors: $u(x) = x^2$ and $v(x) = \ln(x)$. Their derivatives are $u'(x) = 2x$ and $v'(x) = \frac{1}{x}$, using the power rule and the known derivative of $\ln(x)$.

Apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x) = 2x\ln(x) + x^2\cdot\frac{1}{x}$. Simplifying the second term gives $x^2 \cdot \frac{1}{x} = x$, so $f'(x) = 2x\ln(x) + x$.

The term that simplified was the second one, $x^2$ times $\frac{1}{x}$. Multiplying $x^2$ by the derivative of natural log of $x$, which is $\frac{1}{x}$, cancels one factor of $x$, leaving just $x$. The other term, $2x\ln(x)$, involved no such cancellation.