Differentiate for x
(3x^2-4x+3)sin(2x) / 9
(6x-x^-1/2+3)2cos(2x)/9
am i on the right track -
This guy here is the product rule
$\displaystyle \displaystyle y= vu \implies y'=vu'+uv'$
used when you differentiate a product.
In your case make $\displaystyle \displaystyle v= \frac{(3x^2-4x+3)}{9}, u =\sin(2x)$ ifnd $\displaystyle u' $ and $\displaystyle v' $
so we get,
$\displaystyle
\displaystyle \left(\frac{(3x^2-4x+3)\sin(2x)}{9}\right)' = \frac{(3x^2-4x+3)}{9}(\sin(2x))'}+\left(\frac{(3x^2-4x+3)}{9}\right)'\sin(2x)$