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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.

  • 1
    Can you explain in what sense you "need to know algebraically"?2012-10-06
  • 0
    More like I just need to know how to prove it.2012-10-06

4 Answers 4

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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

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

0

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

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.

$$\lim_{h \rightarrow 0} \frac{ \cos(x + h) - \cos(x) }{h}$$

$$\lim_{h \rightarrow 0} \frac{ \cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x) }{h}$$

$$\lim_{h \rightarrow 0} \frac{ \cos(x)\cos(h) - \cos(x) - \sin(x)\sin(h) }{h}$$

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

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

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

$$\cos(x) \lim_{h \rightarrow 0} \frac{ \cos(h) - 1 }{h} - \sin(x) \lim_{h \rightarrow 0} \frac{\sin(h)}{h}$$

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

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

The above then becomes

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

  • 1
    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.2012-10-06
  • 0
    Please learn some latex. http://www.thestudentroom.co.uk/wiki/LaTex. I have done the first few for you.2014-12-13