Derivative of Sine Function

Theorem: If $$y = \sin(x)\,$$, then $$\frac{dy}{dx}=\cos(x)$$ Proof:

$$\frac{dy}{dx}=\frac{\sin(x+h)-\sin(x)}{h}$$

$$\frac{dy}{dx}=\frac{\sin(x)\cos(h)-\sin(h)\cos(x)-\sin(x)}{h}$$

Then since: $$\lim_{h \to 0}\sin(h) = h\mbox{ and }\lim_{h \to 0}\cos(h) = 1$$

We have: $$\frac{dy}{dx}=\frac{\sin(x)1-h\cos(x)-\sin(x)}{h}$$

$$\frac{dy}{dx}=\frac{h\cos(x)}{h}$$

$$\frac{dy}{dx}=\cos(x)$$