Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 5 down vote accepted

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 $\cos h+1$.

share|improve this answer
    
wolframalpha.com/input/… Why doe this state that lim h->0 cos(h-1)/h != 0? –  Alex Jan 5 at 22:55
    
@aix $\frac{\cos(h-1)}h\ne\frac{\cos h-1}h$ (not to mention another syntactic error you submitted there) –  Hagen von Eitzen Jan 5 at 22:58
    
Oh I see--silly me. –  Alex Jan 5 at 23:18

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...

share|improve this answer
    
I don't understand this answer. Could you possibly break this down into more steps? –  Alex Jan 5 at 23:59
    
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. –  DonAntonio Jan 6 at 0:28
    
that's it--as in the other answer, explaining or naming identities used is very helpful. Thank you for the help. –  Alex Jan 6 at 4:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.