## Calculus (3rd Edition)

$$y'= -\cos x\sin(\sin x)\cos(\cos(\sin x)).$$
Since $y=\sin(\cos(\sin x))$, then by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and using that $(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$, the derivative $y'$ is given by $$y'=\cos(\cos(\sin x))(-\sin(\sin x)(\cos x))\\=-\cos x\sin(\sin x)\cos(\cos(\sin x)).$$