Question

What is the derivative of 2sinx and 2cos2x? i need to find the rules in finding derivative of such functions.

Answer 1

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:

$y=2\cos { 2x } =2u\\ \\ \frac { dy }{ du } =2\\ \\ u=\cos { 2x } =\cos { p } \\ \\ \frac { du }{ dp } =-\sin { p=-\sin { 2x } } \\ \\ p=2x\\ \\ \frac { dp }{ dx } =2\\ \\ \frac { dp }{ dx } \cdot \frac { du }{ dp } =-2\sin { 2x } =\frac { du }{ dx } \\ \\ \therefore \quad \frac { dy }{ du } \cdot \frac { du }{ dx } =-4\sin { 2x } =\frac { dy }{ dx }$