One must use the chain rule to solve such problems:
\(\frac { dy }{ du } \cdot \frac { du }{ dx } =\frac { dy }{ dx }\)
One must also know that:
\(If\quad y=\sin { x } ,\quad \frac { dy }{ dx } =\cos { x } \\ \\ If\quad y=\cos { x } ,\quad \frac { dy }{ dx } =-\sin { x }\)
This means that:
\(y=2\sin { x } =2u\\ \\ \frac { dy }{ du } =2\\ \\ u=\sin { x } \\ \\ \frac { du }{ dx } =\cos { x } \\ \\ \therefore \quad \frac { dy }{ du } \cdot \frac { du }{ dx } =2\cos { x } =\frac { dy }{ dx }\)
And also:
\(y=2\cos { 2x } =2u\\ \\ \frac { dy }{ du } =2\\ \\ u=\cos { 2x } =\cos { p } \\ \\ \frac { du }{ dp } =-\sin { p=-\sin { 2x } } \\ \\ p=2x\\ \\ \frac { dp }{ dx } =2\\ \\ \frac { dp }{ dx } \cdot \frac { du }{ dp } =-2\sin { 2x } =\frac { du }{ dx } \\ \\ \therefore \quad \frac { dy }{ du } \cdot \frac { du }{ dx } =-4\sin { 2x } =\frac { dy }{ dx } \\\)