(ii) The Derivative of the Cosine at $x=0$: $\cos'(0)=0$.
Proof: Recall that cos(0) = 1, so the difference quotient simplifies
to $\frac {\cos(0+h) - \cos(0)} h = \frac {\cos(h) - 1}
h$ (8)
Now we multiply the numerator and denominator of this expression by $\cos(h) + 1$ to obtain
$$\frac {\cos(0+h) - \cos(0)} h = \frac {\cos(h) - 1}
h = \frac {(cos(h) + 1)(\cos(h) - 1)}{(cos(h) + 1) h} =\frac {\cos^2(h) -
1}{(cos(h) + 1) h}=\frac {-\sin^2(h)}{(cos(h) + 1) h} = -\frac
{\sin(h)} h \cdot \frac {\sin(h)}{(cos(h) + 1)}$$ (9)
Now it's time to thimk! (Step IV)
When $h \to 0$, we saw in (7) that $ \frac {\sin(h)} h
\to 1$, while $\sin(h)\to 0$ and $\cos(h)\to 1$, so $\cos(h)+ 1 \to 2$.
Putting this all together it should make sense now that
$$\cos'(0) = \lim_{h \to 0} \frac {\cos(h)-1} h =
\lim_{h \to 0^-}\frac {\sin(h)} h \cdot \frac {\sin(h)}{(cos(h) + 1)} =0
$$ (10)
EOP.