Proving that $f(x) = \cos(x)\implies f'(x) = -\sin(x)$ using the definition of a derivative

Question

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.

Answer 1

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 $\cos h+1$.

Answer 2

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...