Answer
$y' = 3x^2\ln x + x^2$
Work Step by Step
In order to derivate this function you have to apply the product rule
$\dfrac{d}{dx}(ab)= a'b+ab'$
Identify the functions a and b, and derivate them
$a=x^3$
$a'=3x^2$
$b=\ln x$
$b'=\dfrac{1}{x}$
Then undo the substitution, simplify and get the answer:
$y' = (3x^2)(\ln x)+(x^3)(\dfrac{1}{x})$
$y' = 3x^2\ln x + x^2$
