## Calculus (3rd Edition)

$$y'= -4(3x^2-\sin x) (x^3+\cos x)^{-5}.$$
Since $y=(x^3+\cos x)^{-4}$, then by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ the derivative $y'$ is given by: $$y'=-4 (x^3+\cos x)^{-5}(x^3+\cos x)'\\=-4 (x^3+\cos x)^{-5}(3x^2-\sin x)\\=-4(3x^2-\sin x) (x^3+\cos x)^{-5}.$$ Here, we used the fact that $(\cos x)'=-\sin x$.