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!
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$\begingroup$Hint
Chain Rule
$$h(g(x))\to h'(g(x))\cdot g'(x)$$The chain rule is used for an inner function which in your case would be $2x$, and the outer would be $\cos(u)$, $u=2x$. So then take the derivative of the outer first giving you the result below
$$\cos(2x)\to -\sin(2x) \cdot2$$
Trigonometric Identities
$$\sin(2x)=2\sin(x)\cos(x)$$
Derivative Trigonometric Identities
$\sin(u)\to u'\cos(u)\\\\\\\\\\\\\\\\\cos(u)\to -u'\sin(u)$
Final Answer$$f(x)=\cos(2x)-2\sin(x)$$$$f'(x)=(2)(-\sin(2x))-2(\cos(x))$$$$f'(x)=-2\sin(2x)-2\cos(x)$$$$f'(x)=-2(2\sin(x)\cos(x))-2\cos(x)$$$$f'(x)=-2\cos(x)(1+2\sin(x))$$
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 4 $\begingroup$$\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.
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