Answer
$$ y'= \cos(1-3x)+3x\sin(1-3x).$$
Work Step by Step
Since $ y=x\cos(1-3x)$, then by the chain and product rules, the derivative $ y'$ is given by
$y'= \cos(1-3x) -x\sin(1-3x)(-3)$
$=\cos(1-3x)+3x\sin(1-3x)$
(Recall that $(\cos x)'=-\sin x$.)
