Finding the derivative of $\cos 2 x - 2 \sin x$ So, I've been teaching myself calculus, and I'm very new to all of this, so apologies in advance for what is probably a rather dumb question. 
I'm trying to find the derivative of the function $f(x) = \cos 2x - 2 \sin x$.
I'm 99% sure that the derivative of $\cos$ is $-\sin$, and that the derivative of $\sin$ is $\cos$. So I got $-\sin 2 + \cos 1$. I just moved through from left to right - $2x$ becomes $2$ and $+-2$ becomes just plus the next thing because constants disappear, etc.
However, the answer in my book is $-2 \cos x(1+2 \sin x)$. I have no clue how the book got this. Just in case I misunderstood the problem, it says,

In Exercises 1 through 14, determine the derivative $f'(x)$. In each case it is understood that $x$ is restricted to those values for which the formula for $f(x)$ is meaningful.

And then for each problem it gives a function like this particular one.
Any help would be appreciated. Thanks!

So, the trick to this one is trigonometric identities:
$$\sin(2x)=2\sin(x)\cos(x)$$
Basically, we end up with
$$-2\cos(x)(1+2\sin(x))=-2\cos(x)-2\underbrace{(2\sin(x)\cos(x))}_{\large\sin(2x)}$$
And notice a small chain rule in the initial problem.
