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Find $\dfrac{d}{dx}(\cos x)$

I know the answer is $-\sin x$ only by process of elimination. I can find solution graphically but I need to know algebraically. Here is my proof so far.
$\begin{align*} \dfrac{d}{dx}\cos x=\lim_{h\to 0}\dfrac{\cos (x+h)-\cos x}{h} &=\lim_{h\to 0}\dfrac{\cos x\cos h-\sin x\sin h-\cos x}{h} \end{align*}$

And that's where I end up and I have no clue where to go from here. Can someone please give me the next step but not the complete answer.

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Can you explain in what sense you "need to know algebraically"? –  Hurkyl Oct 6 '12 at 3:45
    
More like I just need to know how to prove it. –  Austin Broussard Oct 6 '12 at 16:30

4 Answers 4

up vote 4 down vote accepted

You need to know how to evaluate the following limits: $$ \lim_{h \rightarrow 0} \frac{ \cos{h} - 1}{h}, \quad \lim_{h \rightarrow 0} \frac{ \sin{h}}{h}$$

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If you know $d \sin x / dx$, then relate $\cos$ to $\sin$.

Or find some other way to combine things to relate what you know to what you don't know (or what you don't know to itself). One mildly amusing approach is

$$ \frac{d}{dx} \left( \sin^2 x + \cos^2 x \right) $$

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$$\cos h -1=\frac{(\cos h -1)(\cos h+1)}{\cos h +1}$$

Alternately, use the fact that $$\cos 2t=\cos^2 t-\sin^2 t=2\cos^2 t-1=1-2\sin^2 t.$$

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Sorry I cant I dont know how to write math symbols here hope u understand like that.

$\lim_{h \rightarrow 0} [ cos(x + h) - cos(x) ] / h$

$\lim_{h \rightarrow 0} [ cos(x)cos(h) - sin(x)sin(h) - cos(x) ] / h$

$\lim_{h \rightarrow 0} [ cos(x)cos(h) - cos(x) - sin(x)sin(h) ] / h$

$\lim_{h \rightarrow 0} [ cos(x) [ cos(h) - 1 ] - sin(x)sin(h) ] / h$

$\lim_{h \rightarrow 0} [ cos(x) [ cos(h) - 1 ]/h - [sin(x)sin(h)]/h$

$\lim_{h \rightarrow 0} [ cos(x) [cos(h) - 1]/h ] - \lim_{h \rightarrow 0} sin(x)sin(h)/h$

Factor $cos(x)$ from the first limit, and $sin(x)$ from the second limit.

$\cos(x) lim_{h \rightarrow 0} [ cos(h) - 1 ]/h - sin(x) \lim_{h \rightarrow 0} sin(h)/h$

We need foreknowledge of trig limits to realize that $\lim_{h \rightarrow 0} [ cos(h) - 1 ]/h = 0$, and

$\lim_{h \rightarrow 0} sin(h)/h = 1$

The above then becomes

$cos(x) (0) - sin(x)(1)$ $0 - sin(x)$

     or

$\cos x \cos h - \sin x \sin h - \cos x)/h$ As h is tending to zero..... cos h =1 =(cos x - cos x -sin xsin h)/h = -sin x(sin h)/h sin h/h=1 = - sin x

-sin(x)

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IMO, simply writing a complete answer is usually a poor way to help someone learn to solve problems. Furthermore, the questioner specifically asked not to be given one. –  Hurkyl Oct 6 '12 at 19:24
    
Please learn some latex. thestudentroom.co.uk/wiki/LaTex. I have done the first few for you. –  1234 Dec 13 at 22:20

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