# How do you differentiate f(x) = x² ln(x) ?

Oct 9, 2015

Using the product rule,

${f}^{'} \left(x\right) = x \left(2 \ln \left(x\right) + 1\right)$

#### Explanation:

We have that for every $f \left(x\right)$ such that $f \left(x\right) = g \left(x\right) h \left(x\right)$, ${f}^{'} \left(x\right) = {g}^{'} \left(x\right) h \left(x\right) + g \left(x\right) {h}^{'} \left(x\right)$,

We know that ${\left({x}^{2}\right)}^{'} = 2 x$ and that ${\left(\ln \left(x\right)\right)}^{'} = \frac{1}{x}$, so we just evaluate it

${f}^{'} \left(x\right) = 2 x \cdot \ln \left(x\right) + {x}^{2} \cdot \frac{1}{x} = 2 x \cdot \ln \left(x\right) + x$

Or, putting $x$ in evidence

${f}^{'} \left(x\right) = x \left(2 \ln \left(x\right) + 1\right)$