Answer
$$ y'
=-27\sin x(1+\cos^2x)(\cos^3x+3\cos x+7)^8.$$
Work Step by Step
Recall that $(\cos x)'=-\sin x$.
Recall that $(x^n)'=nx^{n-1}$
Since $ y=(\cos^3x+3\cos x+7)^9$, by using the chain rule, the derivative $ y'$ is given by
$$ y'=9(\cos^3x+3\cos x+7)^8(3\cos^2x(-\sin x) -3\sin x)\\
=-27\sin x(1+\cos^2x)(\cos^3x+3\cos x+7)^8.$$
