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I'm having trouble grasping the concept which proves that the derivative of $f(x) = \cos(x)$ is $f'(x) = -\sin(x)$. It needs to be proven using the definition of a derivative--and I can't quite piece it together in my head. Could somebody clarify the concept for me? I appreciate it very much. I'd just like to re-iterate that it needs to be proven using the definition of a derivative. Thanks again.

I'd appreciate as many steps as possible--also, if you could define some of the laws you are using, that'd be very helpful. My trigonometry skills are very mediocre. I appreciate the help you all are giving, but I'm still very cloudy on my understanding.

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Hint: We want to find $$\lim_{h\to 0} \frac{\cos(x+h)-\cos x}{h}.$$ By the Addition Law for the cosine, we have $\cos(x+h)=\cos x\cos h-\sin x\sin h$. So we want $$\lim_{h\to 0} \frac{\cos x\cos h-\cos x -\sin x\sin h}{h}.$$ Now use the fact that $\lim_{h\to 0} \frac{\sin h}{h}=1$ and $\lim_{h\to 0} \frac{\cos h-1}{h}=0$.

If you know that $\lim_{h\to 0}\frac{\sin h}{h}=1$, you can find $\lim_{h\to 0}\frac{\cos h -1}{h}$ by multiplying top and bottom by $\cos h+1$.

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  • $\begingroup$ wolframalpha.com/input/… Why doe this state that lim h->0 cos(h-1)/h != 0? $\endgroup$ – alxmke Jan 5 '14 at 22:55
  • $\begingroup$ @aix $\frac{\cos(h-1)}h\ne\frac{\cos h-1}h$ (not to mention another syntactic error you submitted there) $\endgroup$ – Hagen von Eitzen Jan 5 '14 at 22:58
  • $\begingroup$ Oh I see--silly me. $\endgroup$ – alxmke Jan 5 '14 at 23:18
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With a little trigonometry:

$$(\cos x_0)'=\lim_{x\to x_0}\frac{\cos x-\cos x_0}{x-x_0}=\lim_{x\to x_0}-\frac{2\sin\frac{x+x_0}2\sin\frac{x-x_0}2}{x-x_0}=$$

$$=\;-\lim_{x\to x_0}\sin\frac{x+x_0}2\lim_{x\to x_0}\frac{\sin\frac{x-x_0}2}{\frac{x-x_0}2}=\;-\sin x_0\cdot 1=-\sin x_0$$

Note the splitting of the limit in the product of two limits is justified since each of those two limits exists finitely...

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  • $\begingroup$ I don't understand this answer. Could you possibly break this down into more steps? $\endgroup$ – alxmke Jan 5 '14 at 23:59
  • $\begingroup$ What is what you don't understand? Perhaps that $$\cos a-\cos b=-2\sin\left(\frac{a+b}2\right)\sin\left(\frac{a-b}2\right)\;?$$ It is a trigonometric identity. Google it. $\endgroup$ – DonAntonio Jan 6 '14 at 0:28
  • $\begingroup$ that's it--as in the other answer, explaining or naming identities used is very helpful. Thank you for the help. $\endgroup$ – alxmke Jan 6 '14 at 4:03

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