Answer:
[tex]\large\boxed{\dfrac{dy}{dx}=2\sin(3x)+6\cos(3x)}[/tex]
Step-by-step explanation:
[tex]\dfrac{dy}{dx}=(2x\sin(3x))'\\\\\text{use}\ \bigg(g(x)\cdot f(x)\bigg)'=g'(c)f(x)+g(x)f'(x)\\\\\text{and}\ \bigg(f(g(x))\bigg)'=f'(g(x))\cdot g'(x)\\\\\dfrac{dy}{dx}=(2x)'(\sin(3x))+(2x)(\sin(3x))'\\\\\dfrac{dy}{dx}=2\sin(3x)+(2x)(\cos(3x)\cdot(3x)')\\\\\dfrac{dy}{dx}=2\sin(3x)+2x\cos(3x)\cdot3\\\\\dfrac{dy}{dx}=2\sin(3x)+6\cos(3x)[/tex]
