## Calculus (3rd Edition)

$$y''=32 \cos (8x+18).$$
Recall that $(\sin x)'=\cos x$. Given $y=\sin^2 (4x+9)$, then, by using the chain rule, we have $$y'=2\sin(4x+9) \cos (4x+9) (4)=8\sin(4x+9) \cos (4x+9) .$$ One can simplify $y'$ using the fact that $2\sin x \cos x=\sin2x$ as follows $$y'=4\sin (8x+18)$$ and, by using the chain rule, the second derivative $y''$ is given by $$y''=4\cos (8x+18) (8)=32 \cos (8x+18).$$