Compute $\cos'(0)$ I know that $\cos'(0)$ is $0$, but my work follows:
$$\begin{align}\cos'(0) &= \lim_{\Delta x\to 0}\frac{f(0+\Delta x) - f(0)}{\Delta x}\\
&= \frac{\cos(0+\Delta x) - \cos(0)}{\Delta x}\\
&= \frac{\cos(0) - \cos(0)}{0} = \frac{0}{0}\end{align}$$
Where have I gone wrong? What can I do to show that it equals 0?
(Note: No derivative rules are allowed in my calculus class yet, just difference quotients)
 A: You can't just substitute in 0 for the limit - that loses all important information and leaves you with an indeterminate form. You must be wittier - usually, with $\cos$ and $\sin$ and their derivatives (even with difference quotients), one ends up using the following inequalities.
$$\cos A - \cos B = -2 \sin {\frac{1}{2} (A + B)} \cos {\frac{1}{2} (A + B)}$$
$$\cos(x) = \sin(\frac{\pi}{2} - x)$$
Or you could use the difference of $\sin$ angles too, depending on how you tackle the proof. The key is to not get rid of the $\Delta x$ too early - evaluate the limit only when you come across a form that you know how to evaluate.
I think that gives you a next step, right?
A: Because it is easier to type, I will write $h$ instead of $\Delta x$.  We want to find
$$\lim_{h \to 0} \frac{\cos h -1}{h}.$$
Multiply "top" and "bottom" by $\cos h+1$. We want 
$$\lim_{h \to 0} \frac{(\cos h -1)(\cos h+1)}{h(\cos h+1)}.$$
On top we now have $-\cos^2 h +1$, that is, $-\sin^2 h$. So we want
$$\lim_{h \to 0} \frac{-\sin^2 h}{h(\cos h+1)}\qquad\text{that is,}\qquad \lim_{h \to 0} \left(\frac{\sin h}{h}\cdot \frac{-\sin h}{\cos h+1}\right).$$
Finally, let $h \to 0$. We are allowed to use the fact that $\lim_{h\to 0}\frac{\sin h}{h}=1$. And it is clear that $\lim_{h\to 0}\frac{-\sin h}{\cos h +1}=0$.  So our limit is $0$.
Comment: The idea used above is related to the process of "rationalizing the numerator," which you may have seen already, for example in finding the derivative of $\sqrt{x}$ at $x=3$ from the definition of the derivative. 
