Originally Posted by

**topsquark** First the product rule, then the chain rule.

$\displaystyle f(x) = x \cdot ln^2(x)$

$\displaystyle f^{\prime}(x) = (x)^{\prime} \cdot (ln^2(x)) + x \cdot (ln^2(x))^{\prime}$

Now,

$\displaystyle (ln^2(x))^{\prime} = 2 \cdot ln(x) \cdot \frac{1}{x}$, so

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

$\displaystyle f^{\prime}(x) = ln^2(x) + 2ln(x)$

Then

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

$\displaystyle f^{\prime \prime}(x) = \frac{2ln(x)}{x} + \frac{2}{x}$

-Dan