Find $y'$ then sub $x=2$
Use product rule: If $y=u(x)\times v(x)$ where $u,v$ are functions of $x$, then
$$y'=uv'+u'v$$
Here $$u(x)=(3x-1)^3, v(x)=(2x-1)^2$$$y'=(3x-1)^3\times 2(2x-1)\times 2\\+3(3x-1)^2\times 3(2x-1)^2$
$y'=(3(2)-1)^3\times 2(2(2)-1)\times 2\\+3(3(2)-1)^2\times 3(2(2)-1)^2$
$y'=3525$
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