Derivative of Sine Function

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

$$\frac{dy}{dx}=\lim_{h \to 0}\frac{\sin(x+h)-\sin(x)}{h}$$

$$\frac{dy}{dx}=\lim_{h \to 0}\frac{\sin(x)\cos(h)+\sin(h)\cos(x)-\sin(x)}{h}$$

Collecting terms containing $$\sin(x)$$ and factoring yields:

$$\frac{dy}{dx}=\lim_{h \to 0}\frac{\sin(x)(\cos(h)-1)+\sin(h)\cos(x)}{h}$$

Separating the fraction, we have:

$$\frac{dy}{dx}=\lim_{h \to 0}\frac{\sin(x)(\cos(h)-1)}{h}+\lim_{h \to 0}\frac{\sin(h)\cos(x)}{h}$$

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

We have: $$\frac{dy}{dx}=\sin(x)(0) + (1)\cos(x)$$

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