Derivative of a polynomial of sines:

Via the Scalar Multiple rule, we can treat this as .

We can go a step further and consider it as involving powers and apply the Power and Chain rules:

, where

Using the derivative of the sine, we also have

Combining these results, we have

Finally, we substitute for u and simplify:

Note that the argument of the sines and cosines in the result is the same as the argument of the sine in the original problem.