The derivative of $\sin x$ is $\boxed{\cos x}$. We will use the limit definition of derivative to prove this.
\[ \begin{align*} \frac{d}{dx} f(x) &= \lim_{h \to 0} \dfrac{f(x + h) – f(x)}h \end{align*} \]

We will also use the trigonometric identity $\sin(a + b) = \cos a \sin b + \sin a \cos b$ (line 2) and $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1$ (line 4) to derive this:

\[ \begin{align*} \frac{d}{dx} \sin x &= \lim_{h \to 0} \dfrac{\sin(x + h) – \sin x}h\\ &= \lim_{h \to 0} \dfrac{\cos x \sin h + \sin x \cos h – \sin x}h\\ &= \lim_{h \to 0} \dfrac{\cos x \sin h}{h} + \dfrac{\sin x \cos h – \sin x}{h}\\ &= \cos x \cdot \lim_{h \to 0} \dfrac{\sin h}{h} + \lim_{h \to 0} \dfrac{\sin x(\cos h – 1)}{h}\\ &= \cos x + \lim_{h \to 0} \sin x \cdot \lim_{h \to 0} \dfrac{\cos h – 1}{h}\\ \end{align*} \]
Evaluating $\displaystyle \lim_{h \to 0} \dfrac{\cos h – 1}{h}$ by using $\sin^2 x + \cos^2 x = 1$:
\[ \begin{align*} \lim_{h \to 0} \dfrac{\cos h – 1}{h} &= \lim_{h \to 0} \dfrac{\cos h – 1}{h} \cdot \dfrac{\cos h + 1}{\cos h + 1}\\ &= \lim_{h \to 0} \dfrac{\cos^2 h – 1}{h(\cos h + 1)}\\ &= \lim_{h \to 0} \dfrac{-\sin^2 h}{h(\cos h + 1)}\\ &= (-1)\lim_{h \to 0} \left(\dfrac{\sin h}{h}\right)(\sin h)\left(\dfrac1{\cos h + 1}\right)\\ &= (-1)(1)(0)\left(\dfrac12\right) = 0 \end{align*} \]
We get the end result that $\boxed{\frac{d}{dx} \sin x = \cos x + (1)(0) = \cos x}$.