How do you find the derivative of ##y=sin2x-2sinx##?

By taking the derivative of both sides, you get:

##y' = 2cos2x - 2cosx##

To understand this, you must know for starters, the derivative of the basic trigonometric functions.

##dy/dx sinx = cosx## ##dy/dx cosx = -sinx##

If x happens to be a function (in your case 2x), then you must multiply it by the derivative of the function inside (derivative of 2x is 2).

##dy/dx sinu = cosu*u'## ##dy/dx cosu = -sinu*u'##

By taking the derivative of sin2x, we firstly get cos2x then we multiply it by the derivative of u, which comes out to be 2cos2x. The rest is simple. Recall that the derivative of sinx is cosx. Just apply that here and multiply that by your -2 in front.

If all goes well, then you should get an answer of

##y' = 2cos2x - 2cosx##