Answer:
[tex]6\, (x^{2}\, \cos(2\, x) + x\, \sin(2\, x))[/tex], or equivalently [tex]6\, (x^{2}\, \cos^{2}(x) - x^{2}\, \sin^{2}(x) + 2\, x \, \sin(x) \, \cos(x))[/tex].
Step-by-step explanation:
Make use of the double-angle identity of [tex]\sin(x)[/tex]:
[tex]\sin(2\, x) = 2\, \sin(x) \, \cos(x)[/tex].
[tex]\begin{aligned} y &= 6\, x^{2}\, \sin(x)\, \cos(x) \\ &= 3\, x^{2}\, (2\, \sin(x)\, \cos(x)) \\ &= 3\, x^{2} \, \sin(2\, x)\end{aligned}[/tex].
Apply the product rule [tex](\ast)[/tex] and the chain rule [tex](\ast \ast)[/tex]:
[tex]\begin{aligned} & \frac{d}{dx}\left[3\, x^{2} \, \sin(2\, x)\right] \\ =\; & 3\, x^{2}\, \frac{d}{dx}[\sin(2\, x)] + \frac{d}{dx}\left[3\, x^{2}\right]\, \sin(2\, x) && (\ast) \\ =\; & 3\, x^{2}\, (2\, \cos(2\, x)) + 6\, x\, \sin(2\, x) && (\ast\ast) \\ =\; & 6\, x^{2}\, \cos(2\, x) + 6\, x\, \sin(2\, x)\end{aligned}[/tex].
Optionally, rewrite [tex]\cos(2\, x)[/tex] and [tex]\sin(2\, x)[/tex] using double-angle identities:
[tex]\begin{aligned} & \frac{d}{dx}\left[3\, x^{2} \, \sin(2\, x)\right] \\ =\; & \cdots \\ =\; & 6\, x^{2}\, \cos(2\, x) + 6\, x\, \sin(2\, x) \\ =\; & 6\, x^{2}\, (\cos^{2}(x) - \sin^{2}(x))+ 6\, x\, (2\, \sin(x)\, \cos(x)) \\ =\; & 6\, x^{2}\, \cos^{2}(x) - 6\, x^{2}\, \sin^{2}(x) + 12\, x\, \sin(x)\, \cos(x)\end{aligned}[/tex].
