## Calculus, 10th Edition (Anton)

$y' = 3x^2\ln x + x^2$
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: $y' = (3x^2)(\ln x)+(x^3)(\dfrac{1}{x})$ $y' = 3x^2\ln x + x^2$