Find $\frac{d}{dx}(\cos x)$ 
Find $\dfrac{d}{dx}(\cos x)$

I know the answer is $-\sin x$ only by process of elimination. I can find solution graphically but I need to know algebraically. Here is my proof so far.
$\begin{align*} \dfrac{d}{dx}\cos x=\lim_{h\to 0}\dfrac{\cos (x+h)-\cos x}{h} &=\lim_{h\to 0}\dfrac{\cos x\cos h-\sin x\sin h-\cos x}{h}
\end{align*}$
And that's where I end up and I have no clue where to go from here. Can someone please give me the next step but not the complete answer.
 A: You need to know how to evaluate the following limits:
$$ \lim_{h \rightarrow 0} \frac{ \cos{h} - 1}{h}, \quad \lim_{h \rightarrow 0} \frac{ \sin{h}}{h}$$
A: If you know $d \sin x / dx$, then relate $\cos$ to $\sin$.
Or find some other way to combine things to relate what you know to what you don't know (or what you don't know to itself). One mildly amusing approach is
$$ \frac{d}{dx} \left( \sin^2 x + \cos^2 x  \right) $$
A: $$\cos h -1=\frac{(\cos h -1)(\cos h+1)}{\cos h +1}$$
Alternately, use the fact that 
$$\cos 2t=\cos^2 t-\sin^2 t=2\cos^2 t-1=1-2\sin^2 t.$$ 
A: Sorry I cant I dont know how to write math symbols here
hope u understand like that.
$$\lim_{h \rightarrow 0} \frac{ \cos(x + h) - \cos(x) }{h}$$
$$\lim_{h \rightarrow 0} \frac{ \cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x) }{h}$$
$$\lim_{h \rightarrow 0} \frac{ \cos(x)\cos(h) - \cos(x) - \sin(x)\sin(h) }{h}$$
$$\lim_{h \rightarrow 0} \frac{ \cos(x) [ \cos(h) - 1 ] - \sin(x)\sin(h) } {h}$$
$$\lim_{h \rightarrow 0} \frac{ \cos(x) [\cos(h) - 1]}{h} - \lim_{h \rightarrow 0} \frac{\sin(x)\sin(h)}{h}$$
Factor $\cos(x)$ from the first limit, and $\sin(x)$ from the second limit.
$$\cos(x) \lim_{h \rightarrow 0} \frac{ \cos(h) - 1 }{h} - \sin(x) \lim_{h \rightarrow 0} \frac{\sin(h)}{h}$$
We need foreknowledge of trig limits to realize that
$$\lim_{h \rightarrow 0} \frac{ \cos(h) - 1 }{h} = 0$$ and
$$\lim_{h \rightarrow 0} \frac{\sin(h)}{h} = 1$$
The above then becomes
$$\cos(x) (0) - \sin(x)(1)$$
$$=0 - \sin(x)$$
$$= -\sin(x)$$
