the derivative of sin x
$${\begin{align}\left[\sin{x}\right]' = \cos{x} \end{align}}$$ or Leibnitz notation using differentials $${\frac{\mathbb{d}\left(\sin{x}\right)}{\mathbb{d}x} = \cos{x} }$$
proof
differential quotient
Applying the differential quotient (using the h-method) where $h = x - x_0$.
$${f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}}$$
to $f(x) = \sin x$ gives us
$${\begin{align}f'(x) & = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h} \newline & = \lim\limits_{h \to 0} \frac{\sin{(x+h)} - \sin{x}}{h} \newline & = \lim\limits_{h \to 0} \frac{\sin{x}\cos{h} + \sin{h}\cos{x} - \sin{x}}{h} \newline & = \lim\limits_{h \to 0} \frac{(\cos{h}-1)\cdot \sin{x} + \sin{h}\cos{x} }{h} \newline & = \lim\limits_{h \to 0} \frac{(\cos{h}-1) }{h} \cdot \sin{x} + \lim\limits_{h \to 0} \frac{\sin{h} }{h} \cdot \cos{x}\end{align}}$$
In order to continue, we need to know what the limits of $\frac{\sin{x}}{x}$ and $\frac{\cos{x}- 1}{x}$ are.
limit of $\frac{\sin{x}}{x}$
area of a sector and of a sector in a unit circle
More on radians and the sector see here
comparing the areas
$$A_{\sin} < A_{sector} < A_{\tan}$$
$$\frac{1}{2}\sin{x}\cdot 1 < \frac{1}{2}x\cdot 1 < \frac{1}{2}\tan x \cdot 1$$
$$ 1 < \frac{x}{\sin x} < \frac{1}{\cos x}$$
$$ 1 > \frac{\sin x}{x} > \cos x$$
The limit of cosine as x approaches 0 is 1.
$${\lim\limits_{x \to 0} \cos{x} = 1}$$
Hence, the limit of $\frac{\sin{x}}{x}$ as it is “squeezed” between limits that are both equal to 1 is, by the squeezing theorem, also equal to 1.
$${\lim\limits_{x \to 0} \frac{\sin{x} }{x} = 1}$$
the limit of $\frac{\cos{x}- 1}{x}$
Using half-angle theorems, it can be shown that
$$\frac{\cos{x}- 1}{x} = - \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \cdot \sin{\frac{x}{2}} $$
Hence, the limit as $x \rightarrow 0$ is
$${\begin{align}\lim\limits_{x \to 0} \frac{\cos{x}- 1}{x} & = - \lim\limits_{x \to 0} \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \cdot \lim\limits_{x \to 0} \sin{\frac{x}{2}} \newline & = -1 \cdot 0 \newline & = 0\end{align}}$$
the derivative of sin x
Using the above results, we can now show that
$${\begin{align}\left[\sin{x}\right]' & = \lim\limits_{h \to 0} \frac{(\cos{h}-1) }{h} \cdot \sin{x} + \lim\limits_{h \to 0} \frac{\sin{h} }{h} \cdot \cos{x} \newline & = 0 \cdot \sin{x} + 1 \cdot \cos{x} \newline & = \cos{x} \end{align}}$$
the derivative of cos x
Similarly, we can show that
$${\begin{align}\left[\cos{x}\right]' = - \sin{x} \end{align}}$$