#### Answer

$$y'=-4\sin(\sec4x)\sec4x\tan4x$$

#### Work Step by Step

$$y'=(\cos(\sec4x))'=-\sin(\sec4x)(\sec4x)'=
-\sin(\sec4x)(\frac{1}{\cos 4x})'=-\sin(\sec4x)(-\frac{1}{\cos^24x})(\cos4x)'=\sin(\sec4x)\frac{1}{\cos^24x}(-\sin4x)(4x)'=
-\sin(\sec4x)\frac{1}{\cos4x}\frac{\sin4x}{\cos4x}\cdot4=
-4\sin(\sec4x)\sec4x\tan4x$$