${x^3}\log x$

Let $y = {x^3}\log x$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = {x^3} \cdot \cfrac{1}{x} + \log x \cdot 3{x^2} = {x^2} + 3{x^2}\log x$

$\Rightarrow$ $\cfrac{{{d^2}y}}{{d{x^2}}} = 2x + 3\left[ {{x^2} \cdot \cfrac{1}{x} + \log x \cdot 2x} \right]$

$= 2x + 3[x + 2x\log x] = 2x + 3x + 6x\log x$
$= 5x + 6x\log x = x(5 + 6\log x)$