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!

$\frac d{dx} \cos x = -\sin x\\
\frac d{dx} \sin x = \cos x$
Those x's are important.  You can't say "the derivative of sin is cos."  "sin" and "cos" have no meaning without the argument $x.$
Next we have the chain rule.  $\cos x$ is a function.  $\cos 2x$ is a composite function.  
$\frac d{dx} f(g(x)) = f'(g(x))g(x)$  or $\frac {df}{dg}\frac {dg}{dx}$ using Leibnitz handy notation.  Now for the disclaimer, $\frac {df}{dg}$ is not a true fraction, but by the notation it behaves like one.  And when Liebnitz thought up the notation he thought it was a true fraction.
$\frac d{dx} \cos 2x = (-\sin 2x)(2) = -2 \sin 2x$
put it all toghether
$\frac d{dx} \cos 2x - 2\sin x= -2\sin 2x - 2 \cos x$
And I would be inclined to leave it here.  The book answer has applied a trig identity $\sin 2x = 2\sin x\cos x$ but I don't think that adds simplicity or clarity.
