با استفاده از فرمول مشتق $fg$ داریم: $y=\sin x\cos 3x$  ${y}'=\cos x.\cos 3x-3\sin 3x.\sin x$  ${y}'\left( \frac{\pi }{4} \right)=\cos \frac{\pi }{4}\cos \frac{3\pi }{4}-3\sin \frac{3\pi }{4}\sin \frac{\pi }{4}$  ${y}'\left( \frac{\pi }{4} \right)=\left( \frac{\sqrt{2}}{2} \right)\left( \frac{-\sqrt{2}}{2} \right)-3\left( \frac{\sqrt{2}}{2} \right)\left( \frac{\sqrt{2}}{2} \right)=\frac{-1}{2}-\frac{3}{2}=-2$